首页 > 代码库 > NYOJ 搜索题目汇总 NYOJ 20、21、27、42、58、82、202、284、325、353、488、491、523、592、722
NYOJ 搜索题目汇总 NYOJ 20、21、27、42、58、82、202、284、325、353、488、491、523、592、722
NYOJ 搜索题目汇总
NYOJ 20
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
int pre[100005];
vector<int>v[100005];//存储每个结点相邻的边
void DFS(int cur)
{
for(int i = 0; i < v[cur].size(); ++i)//边
{
if(pre[v[cur][i]]) continue; //若存在父节点则继续遍历
pre[v[cur][i]] = cur; //相连节点的父节点为cur
DFS(v[cur][i]); //深搜到底，把一条路上父节点全部找出
}
}
int main()
{
int ncase, num, cur, i, x, y;
scanf("%d", &ncase);
while(ncase--)
{
memset(v, 0, sizeof(v));
memset(pre, 0, sizeof(pre));
scanf("%d %d", &num, &cur);
pre[cur] = - 1; //起点没有父节点
for(i = 0; i < num - 1; ++i)
{
scanf("%d %d", &x, &y);
v[x].push_back(y); //x与y相连
v[y].push_back(x); //y与x也肯定相连
}
DFS(cur); //起点开始深搜
for(i = 1; i <= num; ++i)
printf("%d ", pre[i]); //每个节点的父节点都保存在pri数组，输出即可
}
return 0;
}
NYOJ 21
法一：
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;
int total;
int v1, v2, v3;
bool visit[105][105][105]; //状态是否出现
struct state
{
int a, b, c;
int ceng; //最小步数
state(){
a = b = c = ceng = 0;
}
}b, e;
int BFS(state begin)
{
queue<state> q;
while(!q.empty())
q.pop();
q.push(b);
while(!q.empty())
{
state cur = q.front();
q.pop();
visit[cur.a][cur.b][cur.c] = true;
if(cur.a == e.a && cur.b == e.b && cur.c == e.c) //找到
return cur.ceng;
if(cur.a > 0 && cur.b < v2) //v1->v2
{
state temp = cur;
int tt = min(temp.a, v2 - temp.b);
temp.a -= tt;
temp.b += tt;
if(visit[temp.a][temp.b][temp.c] == false)
{
visit[temp.a][temp.b][temp.c] = true;
temp.ceng++;
q.push(temp);
}
}
if(cur.a > 0 && cur.c < v3) //v1->v3
{
state temp = cur;
int tt = min(temp.a, v3 - temp.c);
temp.a -= tt;
temp.c += tt;
if(visit[temp.a][temp.b][temp.c] == false)
{
visit[temp.a][temp.b][temp.c] = true;
temp.ceng++;
q.push(temp);
}
}
if(cur.b > 0 && cur.a < v1) //v2->v1
{
state temp = cur;
int tt = min(temp.b, v1 - temp.a);
temp.a += tt;
temp.b -= tt;
if(visit[temp.a][temp.b][temp.c] == false)
{
visit[temp.a][temp.b][temp.c] = true;
temp.ceng++;
q.push(temp);
}
}
if(cur.b > 0 && cur.c < v3) //v2->v3
{
state temp = cur;
int tt = min(temp.b, v3 - temp.c);
temp.b -= tt;
temp.c += tt;
if(visit[temp.a][temp.b][temp.c] == false)
{
visit[temp.a][temp.b][temp.c] = true;
temp.ceng++;
q.push(temp);
}
}
if(cur.c > 0 && cur.a < v1) //v3->v1
{
state temp = cur;
int tt = min(temp.c, v1 - temp.a);
temp.c -= tt;
temp.a += tt;
if(visit[temp.a][temp.b][temp.c] == false)
{
visit[temp.a][temp.b][temp.c] = true;
temp.ceng++;
q.push(temp);
}
}
if(cur.c > 0 && cur.b < v2) //v3->v2
{
state temp = cur;
int tt = min(temp.c, v2 - temp.b);
temp.c -= tt;
temp.b += tt;
if(visit[temp.a][temp.b][temp.c] == false)
{
visit[temp.a][temp.b][temp.c] = true;
temp.ceng++;
q.push(temp);
}
}
}
return -1; //没有终状态
}
int main()
{
int ncase;
scanf("%d", &ncase);
while(ncase--)
{
total = 0;
memset(visit, false, sizeof(visit));
scanf("%d %d %d", &v1, &v2, &v3);
b.a = v1, b.b = 0, b.c = 0, b.ceng = 0;
scanf("%d %d %d", &e.a, &e.b, &e.c);
if(v1 < e.a + e.b + e.c)
{
printf("-1\n");
continue;
}
else
printf("%d\n", BFS(b));
}
return 0;
}
法二：
#include <stdio.h>
#include <string.h>
#define MOVE(x) (x=(x+1)%1000)
bool state[100][100];
struct Size
{
int e[3];
}queue[1000];
int V[3],E1,E2,E3;
int BFS()
{
if(E1+E2+E3!=V[0]) return -1;
else if(E1<0 || E2<0 || E3<0) return -1;
else if(V[0]==E1 && 0==E2 && 0==E3) return 0;
else memset(state,false,sizeof(state));
int tpe,cnt=0;
int front,rear;
int num,temp;
Size s;
front=rear=-1; s.e[0]=V[0]; s.e[1]=0; s.e[2]=0;
queue[MOVE(rear)]=s; state[s.e[0]][s.e[1]]=true; num=1; temp=0;
while(front!=rear)
{
s=queue[MOVE(front)];
if(--num==0) cnt++;
for(int i=0;i<3;i++)
{
if(s.e[i])
{
for(int j=(i+1)%3;j!=i;j=(j+1)%3)
{
tpe=(s.e[i]>V[j]-s.e[j])?(V[j]-s.e[j]):s.e[i];
s.e[i]-=tpe; s.e[j]+=tpe;
if(s.e[0]==E1 && s.e[1]==E2 && s.e[2]==E3)
return num?cnt+1:cnt;
if(!state[s.e[0]][s.e[1]])
{
queue[MOVE(rear)]=s;
state[s.e[0]][s.e[1]]=true;
temp++;
}
s.e[i]+=tpe; s.e[j]-=tpe;
}
}
}
if(num==0) num=temp,temp=0;
}
return -1;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d %d %d",&V[0],&V[1],&V[2]);
scanf("%d %d %d",&E1,&E2,&E3);
printf("%d\n",BFS());
}
return 0;
} NYOJ 27
#include<iostream>
#include<cstring>
const int N=100;
using namespace std;
int n,m;
int mp[N][N];
int dir[4][2] = {{-1,0},{0,1},{1,0},{0,-1}};
void dfs(int si,int sj)
{
mp[si][sj] = 0;
int di,i,j;
for(di=0;di<4;di++)
{
i = si + dir[di][0];
j = sj + dir[di][1];
if (i>=0 &&i<m && j>=0 && j<n && mp[i][j])
dfs(i,j);
}
}
int main()
{
int t,cnt;
cin>>t;
while(t--)
{
cin>>m>>n;
int i,j;
for(i=0;i<m;i++)
for(j=0;j<n;j++)
cin>>mp[i][j];
cnt = 0;
for(i=0;i<m;i++)
for(j=0;j<n;j++)
if (mp[i][j]==1)
{
cnt++;
dfs(i,j);
}
cout<<cnt<<endl;
}
return 0;
}
NYOJ 42
#include<stdio.h>
#include<string.h>
int map[1050][1050],vis[1050],chu[1050];
int P,Q;
void dfs(int point)
{
int i;
vis[point]=1;//DFS 深度遍历
for(i=1;i<=P;i++)
if(map[point][i]!=0)
{
chu[point]++;
if(vis[i]==0)
dfs(i);
}
}
int main()
{
int i,j,N;
scanf("%d",&N);
while(N--)
{
int a,b;
int all_vis=1;
memset(map,0,sizeof(map));
memset(vis,0,sizeof(vis));
memset(chu,0,sizeof(chu));
scanf("%d %d",&P,&Q);
for(i=0;i<Q;i++)
{
scanf("%d %d",&a,&b);
map[a][b]=1;
map[b][a]=1;
}
dfs(1);
for(i=1;i<P;i++)
{
if(vis[i]==0)
{
all_vis=0;
break;
}
}
if(all_vis==0) printf("No\n");
else
{
j=0;
for(i=1;i<=P;i++)
if(chu[i]%2==1) j+=1;//欧拉图 度为奇数
if(j==0||j==2) printf("Yes\n");
else printf("No\n");
}
}
}NYOJ 58
#include<stdio.h>
int a[9][9]={
1,1,1,1,1,1,1,1,1,
1,0,0,1,0,0,1,0,1,
1,0,0,1,1,0,0,0,1,
1,0,1,0,1,1,0,1,1,
1,0,0,0,0,1,0,0,1,
1,1,0,1,0,1,0,0,1,
1,1,0,1,0,1,0,0,1,
1,1,0,1,0,0,0,0,1,
1,1,1,1,1,1,1,1,1};
int t,x1,x2,y1,y2,m;
int dfs(int x,int y,int s)
{
if(x==x2&&y==y2){if(m>s)m=s;}
else
{
if(!a[x+1][y]) {a[x+1][y]=1;dfs(x+1,y,s+1);a[x+1][y]=0;}
if(!a[x-1][y]) {a[x-1][y]=1;dfs(x-1,y,s+1);a[x-1][y]=0;}
if(!a[x][y+1]) {a[x][y+1]=1;dfs(x,y+1,s+1);a[x][y+1]=0;}
if(!a[x][y-1]) {a[x][y-1]=1;dfs(x,y-1,s+1);a[x][y-1]=0;}
}
}
int main()
{
int s;
scanf("%d",&t);
while(t--)
{
m=81;s=0;
scanf("%d %d %d %d",&x1,&y1,&x2,&y2);
dfs(x1,y1,s);
printf("%d\n",m);
}
return 0;
}NYOJ 82
法一：
#include <iostream>
#include <queue>
#include <cstdio>
#include <cstdlib>
using namespace std;
typedef struct pos
{
int x,y;
} POS;
int d[4][2]={{-1,0},{0,1},{1,0},{0,-1}};
int bfs(POS from);
char maze[20][20];
int key[5];
queue<pos> que;
int row,col;
POS start;
int main()
{
int i,j;
while(scanf("%d%d",&row,&col) && (row + col))
{
for(i = 0 ; i < 5 ; ++i)
key[i] = 0;
for(i = 0 ; i < row ; ++i)
{
scanf("%s",maze[i]);
for(j = 0 ; j < col ; ++j)
{
if(maze[i][j] == 'S')//记录起点
{
start.x = i; start.y = j;
}
if(maze[i][j]>= 'a' && maze[i][j]<= 'e')//统计各个门所需的钥匙数量
++key[maze[i][j]-'a'];
}
}//for(i)
que.push(start);
if(bfs(start)) puts("YES");
else puts("NO");
while(!que.empty()) que.pop();
}
// system("pause");
return 0;
}
int bfs(POS from)
{
POS p,next;
int i,t;
maze[from.x][from.y] = 'X';
while(!que.empty())
{
p = que.front();
que.pop();
for(i = 0 ; i < 4 ; ++i)
{
next.x = p.x + d[i][0];
next.y = p.y + d[i][1];
if(next.x>=0 && next.x < row && next.y>=0 && next.y < col)
{
if(maze[next.x][next.y] == 'G')//找到终点，结束
return true;
if(maze[next.x][next.y] == '.')//可以走 进队列
{
que.push(next);
maze[next.x][next.y] = 'X';
}
else
{
if(maze[next.x][next.y]>='a' && maze[next.x][next.y] <= 'e')
{
que.push(next);
--key[maze[next.x][next.y]-'a'];//记录找到的钥匙数量
maze[next.x][next.y] = '.';
}
else
if(maze[next.x][next.y]>='A' && maze[next.x][next.y] <= 'E')
{
t = maze[next.x][next.y] - 'A';
/*
这个地方相当关键啊！
*/
if(!key[t])//用钥匙把门打开即可
{
que.push(next);
// maze[next.x][next.y] = 'X';
}
else//钥匙不够，就把原来的p进队列，先走其他的路，以寻找其他的钥匙
{
if(!que.empty()) que.push(p);
}
}
}
}//if
}//for()
}//while()
return 0;
}法二：
#include <queue>
#include <stdio.h>
#include <ctype.h>
#include <string.h>
using namespace std;
typedef struct
{
int x,y;
}Ac;
queue <Ac> S,door[5];
int dir[]={0,1,0,-1,1,0,-1,0};
int from_x,from_y,to_x,to_y;
int lock[5],key[5];
int map[25][25];
bool in[25][25];
bool bfs()
{
while(!S.empty())
{
Ac R = S.front();
S.pop();
if(R.x == to_x && R.y == to_y)
return true;
for(int k=0;k<8;)
{
int x = R.x + dir[k++];
int y = R.y + dir[k++];
if(!in[x][y] && map[x][y]) //路或钥匙
{
S.push({x,y});
in[x][y]=1;
if(map[x][y]>0) //钥匙
{
key[map[x][y]-1]++;
map[x][y]=-1;
}
}
}
if(S.empty())
{
memset(in,0,sizeof(in));
int find=0;
for(int i=0;i<5;i++)
{
if(key[i] && key[i]==lock[i])//芝麻开门
{
key[i]=0,find=1;
while(!door[i].empty())
{
R = door[i].front();
int x = R.x;
int y = R.y;
map[x][y]=-1;
door[i].pop();
}
}
}
if(find)
S.push({from_x,from_y});
else
break;
}
}
return false;
}
void clear()
{
while(!S.empty())
S.pop();
for(int i=0;i<5;i++)
while(!door[i].empty())
door[i].pop();
}
int main()
{
int h,w;
while(scanf("%d%d",&h,&w),h||w)
{
getchar();
memset(lock,0,sizeof(lock));
memset(map,0,sizeof(map));
memset(key,0,sizeof(key));
memset(in,0,sizeof(in));
for(int i=1;i<=h;i++)
{
for(int j=1;j<=w;j++)
{
char c = getchar();
switch(c)
{
case 'S':from_x=i;from_y=j;S.push({i,j});break;
case 'G':to_x=i,to_y=j;
case '.':map[i][j]=-1;break;
}
if(islower(c)) //钥匙
{
lock[c-'a']++;
map[i][j] = c-'a'+1;
}
if(isupper(c) && c!='G' && c!='S' && c!='X')//门
{
door[c-'A'].push({i,j});
}
}
getchar();
}
bool yes=bfs();
if(yes)
puts("YES");
else
puts("NO");
clear();
}
return 0;
}NYOJ 202
#include <iostream>
#include <cstdio>
#include <string.h>
using namespace std;
const int N = 15;
int x[N],y[N],z[N],leftp[N],rightp[N];
void midorder(int x)
{
if(x != -1)
{
midorder(leftp[x]);
printf("%d\n",x);
midorder(rightp[x]);
}
}
int main()
{
int numcase;
scanf("%d",&numcase);
for(int k = 1;k <= numcase;++k)
{
int n;
scanf("%d",&n);
for(int i = 0;i < n;++i)
{
scanf("%d%d%d",&x[i],&y[i],&z[i]);
}
for(int i = 0;i < n;++i)
{
leftp[x[i]] = y[i];
rightp[x[i]] = z[i];
}
int m,xx,yy;
scanf("%d",&m);
while(m--)
{
scanf("%d%d",&xx,&yy);
}
midorder(0);
printf("\n");
}
return 0;
}
NYOJ 284
法一：
#include<iostream>
#include<cstdio>
#include<queue>
using namespace std;
typedef struct node
{
int x,y,step;
}node;
bool operator<(const node &a,const node &b)
{
return a.step>b.step;
}
int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
int main()
{
char map[305][305];
int i,j,ans,n,m;
node a,b;
priority_queue<node>q;
while(scanf("%d %d",&n,&m)&&(n||m))
{
a.x=0;
for(i=0;i<n;i++)
{
scanf("%s*c",&map[i]);
for(j=0;j<m;j++)
{
if(map[i][j]=='Y')
{
a.x=i;
a.y=j;
a.step=0;
q.push(a);
}
}
}
bool flag=false;
while(!q.empty())//广搜
{
a=q.top();
q.pop();
for(i=0;i<4;i++)
{
b.x=a.x+dir[i][0];
b.y=a.y+dir[i][1];
if(b.x>=0 && b.x<n && b.y>=0 && b.y<m)
{
if(map[b.x][b.y]=='T')
{
ans=a.step+1;
flag=true;
break;
}
else if(map[b.x][b.y]=='E')
{
b.step=a.step+1;
q.push(b);
map[b.x][b.y]='S';
}
else if(map[b.x][b.y]=='B')
{
b.step=a.step+2;
q.push(b);
map[b.x][b.y]='S';
}
}
}
if(flag)
break;
}
if(flag)
printf("%d\n",ans);
else
printf("-1\n");
while(!q.empty())
{
q.pop();
}
}
return 0;
}法二：
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
int m, n;
struct node{
int x, y;
node(int a, int b){
x = a; y = b;
}
}*xm, *d;
char maze[301][301];
int dist[301][301], steps;
bool vis[301][301];
queue<node*> q;
void PUSH(int a, int b){
if(vis[a][b] || a < 1 || b < 1 || a > m || b > n || maze[a][b] == 'S' || maze[a][b] == 'R') return;
q.push(new node(a, b));
dist[a][b] = steps+1;
vis[a][b] = 1;
}
int bfs(){
memset(vis, 0, sizeof(vis));
memset(dist, 0, sizeof(dist));
node *cur = new node(xm->x, xm->y);
while(!q.empty()) q.pop();
q.push(cur);
vis[xm->x][xm->y] = 1;
while(!q.empty()){
cur = q.front(); q.pop();
steps = dist[cur->x][cur->y];
if(cur->x == d->x && cur->y == d->y)
return dist[d->x][d->y];
if(maze[cur->x][cur->y] == 'B'){
maze[cur->x][cur->y] = 'E';
vis[cur->x][cur->y] = 0;
PUSH(cur->x, cur->y);
}else if(maze[cur->x][cur->y] != 'R' || maze[cur->x][cur->y] != 'S'){
PUSH(cur->x - 1, cur->y);
PUSH(cur->x, cur->y - 1);
PUSH(cur->x, cur->y + 1);
PUSH(cur->x + 1, cur->y);
}
}
return -1;
}
int main(){
int i, j;
while(scanf("%d%d", &m, &n) && m||n){
for(i = 1; i <= m; i++)
for(j = 1; j <= n; j++){
scanf(" %c ", &maze[i][j]);
if(maze[i][j] == 'Y')
xm = new node(i, j);
else if(maze[i][j] == 'T')
d = new node(i, j);
}
printf("%d\n", bfs());
}
return 0;
}priority_queue
#include<iostream>
#include<functional>
#include<queue>
#include<vector>
using namespace std;
//定义比较结构
struct cmp1{
bool operator ()(int &a,int &b){
return a>b;//最小值优先
}
};
struct cmp2{
bool operator ()(int &a,int &b){
return a<b;//最大值优先
}
};
//自定义数据结构
struct number1{
int x;
bool operator < (const number1 &a) const {
return x>a.x;//最小值优先
}
};
struct number2{
int x;
bool operator < (const number2 &a) const {
return x<a.x;//最大值优先
}
};
int a[]={14,10,56,7,83,22,36,91,3,47,72,0};
number1 num1[]={14,10,56,7,83,22,36,91,3,47,72,0};
number2 num2[]={14,10,56,7,83,22,36,91,3,47,72,0};
int main()
{
priority_queue<int>que;//采用默认优先级构造队列
priority_queue<int,vector<int>,cmp1>que1;//最小值优先
priority_queue<int,vector<int>,cmp2>que2;//最大值优先
priority_queue<int,vector<int>,greater<int> >que3;//注意“>>”会被认为错误，
priority_queue<int,vector<int>,less<int> >que4;////最大值优先
priority_queue<number1>que5; //最小优先级队列
priority_queue<number2>que6; //最大优先级队列
int i;
for(i=0;a[i];i++){
que.push(a[i]);
que1.push(a[i]);
que2.push(a[i]);
que3.push(a[i]);
que4.push(a[i]);
}
for(i=0;num1[i].x;i++)
que5.push(num1[i]);
for(i=0;num2[i].x;i++)
que6.push(num2[i]);
printf("采用默认优先关系:/n(priority_queue<int>que;)/n");
printf("Queue 0:/n");
while(!que.empty()){
printf("%3d",que.top());
que.pop();
}
puts("");
puts("");
printf("采用结构体自定义优先级方式一:/n(priority_queue<int,vector<int>,cmp>que;)/n");
printf("Queue 1:/n");
while(!que1.empty()){
printf("%3d",que1.top());
que1.pop();
}
puts("");
printf("Queue 2:/n");
while(!que2.empty()){
printf("%3d",que2.top());
que2.pop();
}
puts("");
puts("");
printf("采用头文件/"functional/"内定义优先级:/n(priority_queue<int,vector<int>,greater<int>/less<int> >que;)/n");
printf("Queue 3:/n");
while(!que3.empty()){
printf("%3d",que3.top());
que3.pop();
}
puts("");
printf("Queue 4:/n");
while(!que4.empty()){
printf("%3d",que4.top());
que4.pop();
}
puts("");
puts("");
printf("采用结构体自定义优先级方式二:/n(priority_queue<number>que)/n");
printf("Queue 5:/n");
while(!que5.empty()){
printf("%3d",que5.top());
que5.pop();
}
puts("");
printf("Queue 6:/n");
while(!que6.empty()){
printf("%3d",que6.top());
que6.pop();
}
puts("");
return 0;
}
/*
运行结果 ：
采用默认优先关系:
(priority_queue<int>que;)
Queue 0:
83 72 56 47 36 22 14 10 7 3
采用结构体自定义优先级方式一:
(priority_queue<int,vector<int>,cmp>que;)
Queue 1:
7 10 14 22 36 47 56 72 83 91
Queue 2:
83 72 56 47 36 22 14 10 7 3
采用头文件"functional"内定义优先级:
(priority_queue<int,vector<int>,greater<int>/less<int> >que;)
Queue 3:
7 10 14 22 36 47 56 72 83 91
Queue 4:
83 72 56 47 36 22 14 10 7 3
采用结构体自定义优先级方式二:
(priority_queue<number>que)
Queue 5:
7 10 14 22 36 47 56 72 83 91
Queue 6:
83 72 56 47 36 22 14 10 7 3
*/法三：
#include <iostream>
#include <queue>
#include <cstdio>
#include <cstring>
using namespace std;
#define max 301
char map[max][max];
int n,m;
bool vist[max][max];
int d[4][2]={{1,0},{0,1},{-1,0},{0,-1}};
struct node
{
int x,y,c;
friend bool operator <(const node &s1,const node &s2)
{
return s1.c>s2.c;
}
};
node s,e;
int bfs()
{
priority_queue < node > q;
memset(vist,0,sizeof(vist));
node now,t;
s.c=0;
q.push(s);
vist[s.x][s.y]=1;
while (!q.empty())
{
now=q.top();
q.pop();
if (now.x==e.x&&now.y==e.y)
{
return now.c;
}
for (int i=0;i<4;i++)
{
t.x=now.x+d[i][0];
t.y=now.y+d[i][1];
if (t.x>=0&&t.x<n&&t.y>=0&&t.y<m&&!vist[t.x][t.y])
{
vist[t.x][t.y]=1;
if (map[t.x][t.y]=='B')
{
t.c=now.c+2;
q.push(t);
}
else
if (map[t.x][t.y]=='E'||map[t.x][t.y]=='T')
{
t.c=now.c+1;
q.push(t);
}
}
}
}
return -1;
}
int main()
{
int i,j;
while (cin>>n>>m,m+n)
{
for (i=0;i<n;i++)
{
for (j=0;j<m;j++)
{
cin>>map[i][j];
if (map[i][j]=='Y')
s.x=i,s.y=j;
if (map[i][j]=='T')
e.x=i,e.y=j;
}
}
int sum=bfs();
cout<<sum<<endl;
}
return 0;
}NYOJ 325
#include <stdio.h>
#define max(a,b) a>b?a:b
int V,ans,n,w[21],sum[21];
void dfs(int i,int cnt)
{
if(i == 0)
{
ans = max(ans,cnt);
return ;
}
if(ans == V || cnt+sum[i] <= ans) //cut
return ;
if(cnt+w[i] <= V)
dfs(i-1,cnt+w[i]);
dfs(i-1,cnt);
}
int main()
{
while(~scanf("%d",&n))
{
ans = 0;
for(int i=1;i<=n;i++)
{
scanf("%d",&w[i]);
sum[i] = sum[i-1] + w[i];
}
V = sum[n]/2;
dfs(n,0);
printf("%d\n",sum[n]-2*ans);
}
return 0;
} NYOJ 353
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int MAXN = 35;
const int INF = 0xfffffff;
struct Node
{
int x, y, z;
int step;
Node()
{
x = y = z = 0;
step = 0;
}
};
Node TargetPos;
char Graph[MAXN][MAXN][MAXN];
int MinTime[MAXN][MAXN][MAXN];
int row, col, level;
int dir[6][3] = {{1, 0, 0}, {0, 1, 0}, {-1, 0, 0}, {0, -1, 0}, {0, 0, 1}, {0, 0, -1}};
bool Is_CanGo(Node CurPos)
{
if(CurPos.x < 0 || CurPos.x >= level || CurPos.y < 0 || CurPos.y >= row || CurPos.z < 0 || CurPos.z >= col || Graph[CurPos.x][CurPos.y][CurPos.z] == '#')
return false;
return true;
}
void BFS(Node S)
{
queue <Node> Que;
Que.push(S);
while(!Que.empty())
{
Node Curpos = Que.front();
Que.pop();
for(int i = 0; i < 6; ++i)
{
Node t;
t.x = Curpos.x + dir[i][0];
t.y = Curpos.y + dir[i][1];
t.z = Curpos.z + dir[i][2];
if(Is_CanGo(t))
{
t.step = Curpos.step + 1;
if(t.step < MinTime[t.x][t.y][t.z])
{
MinTime[t.x][t.y][t.z] = t.step;
Que.push(t);
}
}
}
}
}
int main()
{
while(scanf("%d %d %d", &level, &row, &col) && (level + row + col))
{
Node StartPos;
for(int i = 0; i < level; ++i)
{
for(int j = 0; j < row; ++j)
{
scanf("%s", Graph[i][j]);
for(int z = 0; z < col; ++z)
{
MinTime[i][j][z] = INF;
if(Graph[i][j][z] == 'S')
StartPos.x = i, StartPos.y = j, StartPos.z = z;
else if(Graph[i][j][z] == 'E')
TargetPos.x = i, TargetPos.y = j, TargetPos.z = z;
}
getchar();
}
if(i != level - 1)
getchar();
}
StartPos.step = 0;
BFS( StartPos );
int t;
t = MinTime[TargetPos.x][TargetPos.y][TargetPos.z];
if(t != INF)
printf("Escaped in %d minute(s).\n", t);
else
printf("Trapped!\n");
}
return 0;
}NYOJ 488
#include<stdio.h>
#include<string.h>
int Map[9][9],floag;
int IsCan(int number,int x,int y){//判断Numebr是否满足数独的条件
int i,j;
for(i=0;i<9;i++){
if(Map[i][y]==number){
return 0;
}
}
for(i=0;i<9;i++){
if(Map[x][i]==number){
return 0;
}
}
x=x/3*3,y=y/3*3;
for(i=x;i<x+3;i++)
for(j=y;j<y+3;j++)
if(Map[i][j]==number)
return 0;
return 1;
}
void dfs(int x,int y){
if(floag) return ;//floag用于标记，当所有数据都搜完的时候就给跳出
if(x==9 && y==0){
int i,j;
for(i=0;i<9;i++){
for(j=0;j<9;j++){
printf("%d ",Map[i][j]);
}
printf("\n");
}
floag=1;
return ;
}
if(y==9){//一行结束进入下一行
dfs(x+1,0);
}
if(Map[x][y]!=0){//已经有数就填下一个
dfs(x,y+1);
}
if(Map[x][y]==0){
int i;
for(i=1;i<=9;i++){
if(IsCan(i,x,y)){
Map[x][y]=i;
dfs(x,y+1);
Map[x][y]=0;//回溯清除，当一次搜索结束的时候但是不满足就将之前的结果清除
}
}
}
}
int main(){
int n;
scanf("%d",&n);
while(n--){
int i,j;
for(i=0;i<9;i++){
for(j=0;j<9;j++){
scanf("%d",&Map[i][j]);
}
}
dfs(0,0);
memset(Map,0,sizeof(Map));
floag=0;
}
return 0;
}
NYOJ 491
法一：
#include"iostream"
#include<cstring>
#include<stdio.h>
#include<time.h>
using namespace std;
typedef unsigned char uchar;
//char cc[2]={'+','-'}; //便于输出
int n, //第一行符号总数
half, //全部符号总数一半
counter; //1计数，即 '-' 号计数
char **p; //符号存储空间
long sum; //符合条件的三角形计数
//t，第一行第 t个符号
void Backtrace(int t)
{
int i, j;
if( t > n )
sum++;
else
{
for(i=0; i<2; ++i) //只取 0('+') 或者 1('-')
{
p[1][t] = i; //第一行第 t个符号
counter += i; //'-'号统计
for(j=2; j<=t; ++j) //当第一行符号 >=2时，可以运算出下面行的某些符号（第一行有几个数就可以相应往下计算几行，每次计算过的就不用重复计算）
{
p[j][t-j+1] = p[j-1][t-j+1]^p[j-1][t-j+2];//通过异或运算下行符号
counter += p[j][t-j+1];
}
if( (counter <= half) && ( t*(t+1)/2 - counter <= half) )//若符号统计未超过半数，并且另一种符号也未超过半数
Backtrace(t+1); //在第一行增加下一个符号
//回溯，判断另一种符号情况
for(j=2; j<=t; ++j)
counter -= p[j][t-j+1];
counter -= i;
}
}
}
int main()
{
while(scanf("%d", &n) != EOF)
{
counter = 0;
sum = 0;
half = n*(n+1)/2;
if( half%2 == 0 )//总数须为偶数，若为奇数则无解
{
half /= 2;
p = new char *[n+1];
for(int i=0; i<=n; ++i)
{
p[i] = new char[n+1];
memset(p[i], 0, sizeof(char)*(n+1));
}
Backtrace(1);
}
printf("%d\n", sum);
}
return 0;
}法二：
#include<iostream>
#include<cstring>
#include<cstdio>
#include<ctime>
#include<algorithm>
using namespace std;
#define M 20
int cnt,zcnt,fcnt,res[M];
bool ope[M][M];
void dfs1(int n,int p)
{
if(p == n)
{
zcnt = fcnt = 0;
for(int i = 1;i <= p;++i)//
for(int j = 1;j <= p - i + 1;++j)
{
if(ope[i-1][j]) fcnt++;
else zcnt++;
if(j == p - i + 1) continue;
if(ope[i-1][j] == ope[i-1][j+1]) ope[i][j] = false;
else ope[i][j] = true;
}
if(zcnt == fcnt && zcnt) cnt++;
return ;
}
dfs1(n,p+1);
ope[0][p+1] = true;
dfs1(n,p+1);
ope[0][p+1] = false;
}
void init()
{
for(int i = 1;i < 20;++i)
{
memset(ope,0,sizeof(ope));
cnt = 0;
if((i*(i+1)/2)%2==0) dfs1(i,0);
res[i] = cnt;
}
}
int main()
{
int n;init();
while(~scanf("%d",&n))
{
printf("%d\n",res[n]);
}
return 0;
}
/*
枚举第一行所有的情况，可以用深搜实现，可根据第一行逐行确定出下一行的状态，，然后算出每种情况两种符号的个数，在运算之前可以先进行一下判断，当(n+1)*n/2为奇数时，结果为零
*/
//超时NYOJ 499
法一：
#include<stdio.h>
#include<string.h>
int a[10][10], count, n, m;
int b[10][10];
void DFS(int x, int y)
{
if(x > n || x < 1 || y > m || y < 1 || a[x][y] == 1)
{
return ;
}
if(x == n && y == m)
{
count++;
return ;
}
printf("%d,%d\n",x,y);
a[x][y] = 1;
DFS(x + 1, y);
DFS(x - 1, y);
DFS(x, y + 1);
DFS(x, y - 1);
a[x][y] = 0;
}
int main()
{
int t, i, j;
scanf("%d", &t);
while(t--)
{
count = 0;
memset(b, 0, sizeof(b));
scanf("%d%d", &n, &m);
for(i = 1; i <= n; i++)
{
for(j = 1; j <= m; j++)
{
scanf("%d", &a[i][j]);
}
}
DFS(1, 1);
printf("%d\n", count);
}
return 0;
}
/*0 1 1 1 1 1 1 1
0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 1
1 1 1 1 1 1 1 1*/法二：
#include <stdio.h>
int count;
int dt[8][8];
int r,c;
void init()
{
int i;
for(i = 0 ; i <= r+1; i++)
{
dt[i][0] = 1;
dt[i][c+1] = 1;
}
for(i = 0; i <= c+1; i++)
{
dt[0][i] = 1;
dt[r+1][i] = 1;
}
}
void print()
{
int i,j;
for(i = 0; i <= r+1; i++)
{
for(j = 0; j <= c+1; j++)
printf("%d ",dt[i][j]);
printf("\n");
}
}
void dfs(int x,int y)
{
if(x == r && y == c)
{
count++;
return;
}
dt[x][y] = 1;
if(!dt[x+1][y]) dfs(x+1,y);
if(!dt[x][y+1]) dfs(x,y+1);
if(!dt[x-1][y]) dfs(x-1,y);
if(!dt[x][y-1]) dfs(x,y-1);
dt[x][y] = 0;
}
int main()
{
int i,j,t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&r,&c);
count = 0;
init();
for(i = 0; i < r; i++)
for(j = 0; j < c; j++)
scanf("%d", &dt[i+1][j+1]);
dfs(1,1);
printf("%d\n",count);
}
return 0;
}NYOJ 523
法一：
#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;
int map[55][55][55],f[]={1,-1,0,0,0,0},g[]={0,0,1,-1,0,0},w[]={0,0,0,0,1,-1};
bool vis[55][55][55],flag;
struct Node
{
int x,y,z,step;
};
int bfs(int a,int b,int c,int time)
{
int i,j,k;
int r,s,t;
memset(vis,0,sizeof(vis));
queue <Node> q;
Node temp={0,0,0,time};
vis[0][0][0]=1;
q.push(temp);
while(!q.empty())
{
temp=q.front();
q.pop();
for(i=0;i<6;i++)
{
r=temp.x+f[i];
s=temp.y+g[i];
t=temp.z+w[i];
if(r>=0&&r<a&&s>=0&&s<b&&t>=0&&t<c&&vis[r][s][t]==0)
{
if(r==a-1&&s==b-1&&t==c-1&&map[r][s][t]==0&&temp.step-1>=0)
{
flag=1;
break;
}
else if(map[r][s][t]==0&&temp.step-1)
{
Node next={r,s,t,temp.step-1};
q.push(next);
vis[r][s][t]=1;
}
}
}
if(flag==1||temp.step-1==0)
break;
}
while(!q.empty())
q.pop();
return time-(temp.step-1);
}
int main()
{
int a,b,c;
int i,j,k,T;
int time;
scanf("%d",&T);
while(T--)
{
memset(map,0,sizeof(map));
flag=0;
scanf("%d%d%d%d",&a,&b,&c,&time);
for(k=0;k<a;k++)
{
for(i=0;i<b;i++)
{
for(j=0;j<c;j++)
{
scanf("%d",&map[k][i][j]);
}
}
}
int cnt=bfs(a,b,c,time);
if(flag==1)
printf("%d\n",cnt);
else
printf("-1\n");
}
return 0;
}法二：
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
const int N = 55;
const int dir[6][3] = {{0, 0, 1}, {0, 0, -1}, {-1, 0, 0}, {1, 0, 0}, {0, 1, 0}, {0, -1, 0}};
struct node
{
int x, y, z;
int ceng;
}que[50000];
int maze[N][N][N];
int a, b, c, t;
int num;
int head, tail;
int Scan()
{
int res = 0 , ch;
while( !( ( ch = getchar() ) >= '0' && ch <= '9' ) )
{
if( ch == EOF ) return 1 << 30 ;
}
res = ch - '0' ;
while( ( ch = getchar() ) >= '0' && ch <= '9' )
res = res * 10 + ( ch - '0' ) ;
return res ;
}
int BFS()
{
node cur;
cur.x = 1, cur.y = 1, cur.z = 1, cur.ceng = 0;
maze[cur.x][cur.y][cur.z] = 0;
que[tail++] = cur;
while(head < tail)
{
node temp = que[head++];
if(temp.ceng > t) //无法到达
return -1;
if(temp.x == a && temp.y == b && temp.z == c && temp.ceng <= t)
return temp.ceng;
for(int i = 0; i < 6; ++i)
{
node now;
now.x = temp.x + dir[i][0]; now.y = temp.y + dir[i][1]; now.z = temp.z + dir[i][2];
if(maze[now.x][now.y][now.z])
{
now.ceng = temp.ceng + 1;
que[tail++] = now;
maze[now.x][now.y][now.z] = 0;
}
}
}
return -1;
}
int main()
{
int ncase;
ncase = Scan();
while(ncase--)
{
num = head = tail = 0;
memset(maze, 0, sizeof(maze));
a = Scan(), b = Scan(), c = Scan(), t = Scan();
for(int i = 1; i <= a; ++i)
for(int j = 1; j <= b; ++j)
for(int k = 1; k <= c; ++k)
{
maze[i][j][k] = Scan() ^ 1;
num += maze[i][j][k];
}
if(maze[a][b][c] == 0 || num < a + b + c - 3) //终点无法走
{
printf("-1\n");
continue;
}
printf("%d\n", BFS());
}
return 0;
}NYOJ 592
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
using namespace std;
int map[105][105]={0};
int mm=10000,n=100;
int fir,end,loop[105][105];
int prime[11000]={1,1,0};
int sx[]={0,0,1,-1},zy[]={1,-1,0,0};
void Build_map() //建立地图
{
int i,j,k;
for(i=1;i<=(n+1)/2;i++)
{
for(j=1;j<=n-2*i+2;j++)
map[i][j+i-1]=mm--;
for(j=1;j<=n-2*i;j++)
map[j+i][n-i+1]=mm--;
for(j=n-i+1;j>=i;j--)
map[n-i+1][j]=mm--;
for(j=n-i;j>=i+1;j--)
map[j][i]=mm--;
}
}
void printf_map()//打印地图
{
int i,j;
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
printf("%d ",map[i][j]);
printf("\n");
}
}
void judge_prime() //筛选法打个素数表
{
int i,j;
for(i=2;i<=10000;i++)
{
if(prime[i]==0)
{
for(j=i+i;j<=10000;j+=i)
prime[j]=1;
}
}
}
struct coordinate
{
int x;
int y;
int step;
}t1;
void find() //找到 fir 的坐标，存在 t1 中
{
int i,j;
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
{
if(map[i][j]==fir)
{
t1.x=i;
t1.y=j;
loop[i][j]=1;
t1.step=0;
return;
}
}
}
int bfs()
{
int i,j,step,x,y;
queue<coordinate> Q;
Q.push(t1);
while(!Q.empty())
{
i=Q.front().x;
j=Q.front().y;
step=Q.front().step;
Q.pop();
if(map[i][j]==end)
return step;
for(int a=0;a<4;a++)
{
x=i+sx[a];
y=j+zy[a];
if(map[x][y]!=0&&prime[map[x][y]]==1&&loop[x][y]==0)
{
coordinate t2={x,y,step+1};
Q.push(t2);
loop[x][y]=1;
}
}
}
return 0;
}
int main()
{
int i,j,nn=0;
Build_map();
judge_prime();
//printf_map();
while(~scanf("%d%d",&fir,&end))
{
nn++;
memset(loop,0,sizeof(loop));
find();
int ans=bfs();
if(ans)
printf("Case %d: %d\n",nn,ans);
else
printf("Case %d: impossible\n",nn);
}
}NYOJ 722
#include<stdio.h>
#include<string.h>
int Map[9][9],floag;
int IsCan(int number,int x,int y){//判断Numebr是否满足数独的条件
int i,j;
for(i=0;i<9;i++){
if(Map[i][y]==number){
return 0;
}
}
for(i=0;i<9;i++){
if(Map[x][i]==number){
return 0;
}
}
x=x/3*3,y=y/3*3;
for(i=x;i<x+3;i++)
for(j=y;j<y+3;j++)
if(Map[i][j]==number)
return 0;
return 1;
}
void dfs(int x,int y){
if(floag) return ;//floag用于标记，当所有数据都搜完的时候就给跳出
if(x==9 && y==0){
int i,j;
for(i=0;i<9;i++){
for(j=0;j<9;j++){
printf("%d ",Map[i][j]);
}
printf("\n");
}
floag=1;
return ;
}
if(y==9){//一行结束进入下一行
dfs(x+1,0);
}
if(Map[x][y]!=0){//已经有数就填下一个
dfs(x,y+1);
}
if(Map[x][y]==0){
int i;
for(i=1;i<=9;i++){
if(IsCan(i,x,y)){
Map[x][y]=i;
dfs(x,y+1);
Map[x][y]=0;//回溯清除，当一次搜索结束的时候但是不满足就将之前的结果清除
}
}
}
}
int main(){
int n;
scanf("%d",&n);
while(n--){
int i,j;
for(i=0;i<9;i++){
for(j=0;j<9;j++){
scanf("%d",&Map[i][j]);
}
}
dfs(0,0);
memset(Map,0,sizeof(Map));
floag=0;
}
return 0;
}
NYOJ 搜索题目汇总 NYOJ 20、21、27、42、58、82、202、284、325、353、488、491、523、592、722
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