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[kuangbin带你飞]专题一 简单搜索
一直拖、放了放久、受不了
Description
Input
每组数据的第一行是两个正整数，n k，用一个空格隔开，表示了将在一个n*n的矩阵内描述棋盘，以及摆放棋子的数目。 n <= 8 , k <= n
当为-1 -1时表示输入结束。
随后的n行描述了棋盘的形状：每行有n个字符，其中 # 表示棋盘区域， . 表示空白区域（数据保证不出现多余的空白行或者空白列）。
Output
Sample Input
2 1#..#4 4...#..#..#..#...-1 -1
Sample Output
21
思路：DFS
#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define ll long long#define N 10int n,k;int ans;int col[N];char mpt[N][N];void dfs(int row,int num){ if(num==k) { ans++; return; } if(row>n) return; dfs(row+1,num); for(int j=1;j<=n;j++) { if(mpt[row][j]==‘#‘ && !col[j]) { col[j]=1; dfs(row+1,num+1); col[j]=0; } }}int main(){ while(scanf("%d%d",&n,&k)!=EOF) { if(n==-1 && k==-1) break; ans=0; memset(col,0,sizeof(col)); for(int i=1;i<=n;i++) { scanf("%s",mpt[i]+1); } dfs(1,0); cout<<ans<<endl; } return 0;}
Description
Is an escape possible? If yes, how long will it take?
Input
L is the number of levels making up the dungeon.
R and C are the number of rows and columns making up the plan of each level.
Then there will follow L blocks of R lines each containing C characters. Each character describes one cell of the dungeon. A cell full of rock is indicated by a ‘#‘ and empty cells are represented by a ‘.‘. Your starting position is indicated by ‘S‘ and the exit by the letter ‘E‘. There‘s a single blank line after each level. Input is terminated by three zeroes for L, R and C.
Output
Escaped in x minute(s).
where x is replaced by the shortest time it takes to escape.
If it is not possible to escape, print the line
Trapped!
Sample Input
3 4 5S.....###..##..###.#############.####...###########.#######E1 3 3S###E####0 0 0
Sample Output
Escaped in 11 minute(s).Trapped!
思路：三维BFS
#include <iostream>#include <cstdio>#include <queue>#include <cstring>using namespace std;#define mod 1000000007#define ll long long#define N 50struct node{ int x,y,z,t;};int n,m,k;int x,y,z;int vis[N][N][N];char mpt[N][N][N];int dir[6][3]={1,0,0,-1,0,0,0,1,0,0,-1,0,0,0,1,0,0,-1};bool judge(int x,int y,int z){ if(x<1 || x>n || y<1 || y>m || z<1 || z>k) return 0; if(mpt[x][y][z]==‘#‘) return 0; if(vis[x][y][z]) return 0; return 1;}void bfs(){ queue<node> q; memset(vis,0,sizeof(vis)); node now,next; now.x=x; now.y=y; now.z=z; now.t=0; vis[x][y][z]=1; q.push(now); while(!q.empty()) { now=q.front(); q.pop(); if(mpt[now.x][now.y][now.z]==‘E‘) { printf("Escaped in %d minute(s).\n",now.t); return; } for(int i=0;i<6;i++) { next=now; next.x+=dir[i][0]; next.y+=dir[i][1]; next.z+=dir[i][2]; if(judge(next.x,next.y,next.z)) { next.t++; vis[next.x][next.y][next.z]=1; q.push(next); } } } cout<<"Trapped!\n";}int main(){ while(scanf("%d%d%d",&n,&m,&k),n+m+k) { for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { for(int l=1;l<=k;l++) { scanf(" %c",&mpt[i][j][l]); if(mpt[i][j][l]==‘S‘) { x=i; y=j; z=l; } } } } bfs(); } return 0;}
Description
Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.
* Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute
* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.
If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?
Input
Output
Sample Input
5 17
Sample Output
4
Hint
#include <iostream>#include <cstdio>#include <queue>#include <cstring>using namespace std;#define mod 1000000007#define ll long long#define N 100000int n,m;int vis[N+10];void bfs(int s){ memset(vis,0,sizeof(vis)); queue<pair<int,int>> q; pair<int,int> now,next; now.first=s; now.second=0; vis[s]=1; q.push(now); while(!q.empty()) { now=q.front(); q.pop(); if(now.first==m) { cout<<now.second<<endl; return; } for(int i=1;i<=3;i++) { next=now; next.second++; if(i==1) next.first-=1; else if(i==2) next.first+=1; else if(i==3) next.first*=2; if(next.first>=0 && next.first<=N && !vis[next.first]) { vis[next.first]=1; q.push(next); } } }}int main(){ while(scanf("%d%d",&n,&m)!=EOF) { if(n==m) { cout<<0<<endl; continue; } bfs(n); } return 0;}
Description
Farmer John knows that an intellectually satisfied cow is a happy cow who will give more milk. He has arranged a brainy activity for cows in which they manipulate an M × N grid (1 ≤ M ≤ 15; 1 ≤ N ≤ 15) of square tiles, each of which is colored black on one side and white on the other side.
As one would guess, when a single white tile is flipped, it changes to black; when a single black tile is flipped, it changes to white. The cows are rewarded when they flip the tiles so that each tile has the white side face up. However, the cows have rather large hooves and when they try to flip a certain tile, they also flip all the adjacent tiles (tiles that share a full edge with the flipped tile). Since the flips are tiring, the cows want to minimize the number of flips they have to make.
Help the cows determine the minimum number of flips required, and the locations to flip to achieve that minimum. If there are multiple ways to achieve the task with the minimum amount of flips, return the one with the least lexicographical ordering in the output when considered as a string. If the task is impossible, print one line with the word "IMPOSSIBLE".
Input
Lines 2.. M+1: Line i+1 describes the colors (left to right) of row i of the grid with N space-separated integers which are 1 for black and 0 for white
Output
Sample Input
4 41 0 0 10 1 1 00 1 1 01 0 0 1
Sample Output
0 0 0 01 0 0 11 0 0 10 0 0 0
思路：枚举第一行状态+模拟
#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <queue>#include <string>#include <map>using namespace std;#define INF 0xfffffff#define ll long long#define N 20int n,m;int res;int mpt[N][N]; //原始状态int ans[N][N]; //保存答案int tmp[N][N]; //翻转模拟数组int vis[N][N]; //记录是否翻转void flip(int i,int j){ tmp[i][j]^=1; if(i-1>=0) tmp[i-1][j]^=1; if(i+1<n) tmp[i+1][j]^=1; if(j-1>=0) tmp[i][j-1]^=1; if(j+1<m) tmp[i][j+1]^=1;}int solve(){ for(int i=1;i<n;i++) { for(int j=0;j<m;j++) { if(tmp[i-1][j]) flip(i,j),vis[i][j]=1; } } int time=0; for(int i=0;i<n;i++) { for(int j=0;j<m;j++) { time+=vis[i][j]; if(tmp[i][j]) return INF; } } return time;}int main(){ int i,j,k; while(scanf("%d%d",&n,&m)!=EOF) { res=INF; for(i=0;i<n;i++) { for(j=0;j<m;j++) { scanf("%d",&mpt[i][j]); } } int MAX=1<<m; for(k=0;k<MAX;k++) //枚举第一行的状态 { memset(vis,0,sizeof(vis)); memcpy(tmp,mpt,sizeof(mpt)); for(j=0;j<m;j++) { if(k&(1<<j)) flip(0,j),vis[0][j]=1; } int time=solve(); if(time<res) { res=time; memcpy(ans,vis,sizeof(vis)); } } if(res==INF) cout<<"IMPOSSIBLE\n"; else { for(i=0;i<n;i++) { for(j=0;j<m;j++) { if(j) cout<<‘ ‘; cout<<ans[i][j]; } cout<<endl; } } } return 0;}
Description
Input
Output
Sample Input
26190
Sample Output
10100100100100100100111111111111111111
思路：BFS
#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <queue>#include <string>using namespace std;#define ll long longint n;void bfs(){ queue<ll> q; q.push(1); while(1) { ll sum=q.front(); if(sum%n==0) { cout<<sum<<endl; return; } q.pop(); q.push(10*sum); q.push(10*sum+1); }}int main(){ while(cin>>n,n) { bfs(); } return 1;}
Description
The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on their offices. — It is a matter of security to change such things every now and then, to keep the enemy in the dark.
— But look, I have chosen my number 1033 for good reasons. I am the Prime minister, you know!
— I know, so therefore your new number 8179 is also a prime. You will just have to paste four new digits over the four old ones on your office door.
— No, it’s not that simple. Suppose that I change the first digit to an 8, then the number will read 8033 which is not a prime!
— I see, being the prime minister you cannot stand having a non-prime number on your door even for a few seconds.
— Correct! So I must invent a scheme for going from 1033 to 8179 by a path of prime numbers where only one digit is changed from one prime to the next prime.
Now, the minister of finance, who had been eavesdropping, intervened.
— No unnecessary expenditure, please! I happen to know that the price of a digit is one pound.
— Hmm, in that case I need a computer program to minimize the cost. You don‘t know some very cheap software gurus, do you?
— In fact, I do. You see, there is this programming contest going on... Help the prime minister to find the cheapest prime path between any two given four-digit primes! The first digit must be nonzero, of course. Here is a solution in the case above.
1033The cost of this solution is 6 pounds. Note that the digit 1 which got pasted over in step 2 can not be reused in the last step – a new 1 must be purchased.
1733
3733
3739
3779
8779
8179
Input
Output
Sample Input
31033 81791373 80171033 1033
Sample Output
670
思路：BFS
#include <iostream>#include <cstdio>#include <queue>#include <cstring>using namespace std;#define mod 1000000007#define ll long long#define N 110struct node{ int x,t;};int n,m;int vis[10005];bool judge(int n){ if(vis[n]) return 0; if(n==1) return 0; for(int i=2;i*i<=n;i++) { if(n%i==0) return 0; } return 1;}void bfs(){ int i,j,tmp[5]; queue<node> q; memset(vis,0,sizeof(vis)); node now,next; now.x=n; now.t=0; q.push(now); while(!q.empty()) { now=q.front(); q.pop(); if(now.x==m) { cout<<now.t<<endl; return; } int tp=now.x,k=4; while(tp) { tmp[k--]=tp%10; tp/=10; } for(i=0;i<4;i++) { next=now; next.t++; if(i==0) { for(j=1;j<10;j++) { next.x=j*1000+tmp[2]*100+tmp[3]*10+tmp[4]; if(judge(next.x)) { vis[next.x]=1; q.push(next); } } } else if(i==1) { for(j=0;j<10;j++) { next.x=tmp[1]*1000+j*100+tmp[3]*10+tmp[4]; if(judge(next.x)) { vis[next.x]=1; q.push(next); } } } else if(i==2) { for(j=0;j<10;j++) { next.x=tmp[1]*1000+tmp[2]*100+j*10+tmp[4]; if(judge(next.x)) { vis[next.x]=1; q.push(next); } } } else { for(j=1;j<10;j++) { next.x=tmp[1]*1000+tmp[2]*100+tmp[3]*10+j; if(judge(next.x)) { vis[next.x]=1; q.push(next); } } } } }}int main(){ int T; scanf("%d",&T); while(T--) { scanf("%d%d",&n,&m); bfs(); } return 0;}
Description
A common pastime for poker players at a poker table is to shuffle stacks of chips. Shuffling chips is performed by starting with two stacks of poker chips,S1 and S2, each stack containing C chips. Each stack may contain chips of several different colors.
The actual shuffle operation is performed by interleaving a chip from S1 with a chip from S2 as shown below for C = 5:
The single resultant stack, S12, contains 2 * C chips. The bottommost chip of S12 is the bottommost chip from S2. On top of that chip, is the bottommost chip from S1. The interleaving process continues taking the 2nd chip from the bottom of S2 and placing that on S12, followed by the 2nd chip from the bottom of S1 and so on until the topmost chip from S1 is placed on top of S12.
After the shuffle operation, S12 is split into 2 new stacks by taking the bottommost C chips from S12 to form a new S1 and the topmost C chips from S12to form a new S2. The shuffle operation may then be repeated to form a new S12.
For this problem, you will write a program to determine if a particular resultant stack S12 can be formed by shuffling two stacks some number of times.
Input
The first line of input contains a single integer N, (1 ≤ N ≤ 1000) which is the number of datasets that follow.
Each dataset consists of four lines of input. The first line of a dataset specifies an integer C, (1 ≤ C ≤ 100) which is the number of chips in each initial stack (S1 and S2). The second line of each dataset specifies the colors of each of the C chips in stack S1, starting with the bottommost chip. The third line of each dataset specifies the colors of each of the C chips in stack S2 starting with the bottommost chip. Colors are expressed as a single uppercase letter (Athrough H). There are no blanks or separators between the chip colors. The fourth line of each dataset contains 2 * C uppercase letters (A through H), representing the colors of the desired result of the shuffling of S1 and S2 zero or more times. The bottommost chip’s color is specified first.
Output
Output for each dataset consists of a single line that displays the dataset number (1 though N), a space, and an integer value which is the minimum number of shuffle operations required to get the desired resultant stack. If the desired result can not be reached using the input for the dataset, display the value negative 1 (−1) for the number of shuffle operations.
Sample Input
24AHAHHAHAHHAAAAHH3CDECDEEEDDCC
Sample Output
1 22 -1
思路：模拟+map标记
include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <queue>#include <string>#include <map>using namespace std;#define ll long longint len;int ans;string s,e;string s1,s2;void shuffle(){ s.clear(); for(int i=0;i<len;i++) { s+=s2[i]; s+=s1[i]; }}int solve(){ int res=2; map<string,bool> m; while(m.find(s)==m.end()) //满足条件就表示出现过了 { m[s]=true; s1=s.substr(0,len); s2=s.substr(len,len); shuffle(); if(s==e) return res; res++; } return -1;}int main ( ){ int T,iCase=1; scanf("%d",&T); while(T--) { cin>>len; cin>>s1>>s2>>e; shuffle(); if(s==e) ans=1; else ans=solve(); cout<<iCase++<<" "<<ans<<endl; } return 0;}
Description
You are given two pots, having the volume of A and B liters respectively. The following operations can be performed:
- FILL(i) fill the pot i (1 ≤ i ≤ 2) from the tap;
- DROP(i) empty the pot i to the drain;
- POUR(i,j) pour from pot i to pot j; after this operation either the pot j is full (and there may be some water left in the pot i), or the pot i is empty (and all its contents have been moved to the pot j).
Write a program to find the shortest possible sequence of these operations that will yield exactly C liters of water in one of the pots.
Input
On the first and only line are the numbers A, B, and C. These are all integers in the range from 1 to 100 and C≤max(A,B).
Output
The first line of the output must contain the length of the sequence of operations K. The following K lines must each describe one operation. If there are several sequences of minimal length, output any one of them. If the desired result can’t be achieved, the first and only line of the file must contain the word ‘impossible’.
Sample Input
3 5 4
Sample Output
6FILL(2)POUR(2,1)DROP(1)POUR(2,1)FILL(2)POUR(2,1)
思路：BFS+路径输出
#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <queue>#include <string>using namespace std;#define ll long long#define INF 0x7fffffff#define MOD 1000007#define N 110struct node{ int a,b,t; string s;};int A,B,C;int vis[N][N];void pour(node &t,int k){ if(k==1) { if(B-t.b>=t.a) t.b+=t.a,t.a=0; else t.a-=(B-t.b),t.b=B; } else { if(A-t.a>=t.b) t.a+=t.b,t.b=0; else t.b-=(A-t.a),t.a=A; }}void show(string s){ int len=s.size(); for(int i=0;i<len;i++) { if(s[i]==‘1‘) printf("FILL(1)\n"); else if(s[i]==‘2‘) printf("FILL(2)\n"); else if(s[i]==‘3‘) printf("DROP(1)\n"); else if(s[i]==‘4‘) printf("DROP(2)\n"); else if(s[i]==‘5‘) printf("POUR(1,2)\n"); else if(s[i]==‘6‘) printf("POUR(2,1)\n"); }}void bfs(){ node now,next; queue<node> q; memset(vis,0,sizeof(vis)); now.a=0; now.b=0; now.t=0; now.s=""; vis[0][0]=1; q.push(now); while(!q.empty()) { now=q.front(); q.pop(); if(now.a==C || now.b==C) { printf("%d\n",now.t); show(now.s); return; } for(int i=1;i<=6;i++) { next=now; next.t++; if(i==1) next.a=A,next.s+=‘1‘; else if(i==2) next.b=B,next.s+=‘2‘; else if(i==3) next.a=0,next.s+=‘3‘; else if(i==4) next.b=0,next.s+=‘4‘; else if(i==5) pour(next,1),next.s+=‘5‘; else if(i==6) pour(next,2),next.s+=‘6‘; if(!vis[next.a][next.b]) { q.push(next); vis[next.a][next.b]=1; } } } printf("impossible\n");}int main(){ while(scanf("%d%d%d",&A,&B,&C)!=EOF) { bfs(); } return 0;}
Description
Fat brother and Maze are playing a kind of special (hentai) game on an N*M board (N rows, M columns). At the beginning, each grid of this board is consisting of grass or just empty and then they start to fire all the grass. Firstly they choose two grids which are consisting of grass and set fire. As we all know, the fire can spread among the grass. If the grid (x, y) is firing at time t, the grid which is adjacent to this grid will fire at time t+1 which refers to the grid (x+1, y), (x-1, y), (x, y+1), (x, y-1). This process ends when no new grid get fire. If then all the grid which are consisting of grass is get fired, Fat brother and Maze will stand in the middle of the grid and playing a MORE special (hentai) game. (Maybe it’s the OOXX game which decrypted in the last problem, who knows.)
You can assume that the grass in the board would never burn out and the empty grid would never get fire.
Note that the two grids they choose can be the same.
Input
The first line of the date is an integer T, which is the number of the text cases.
Then T cases follow, each case contains two integers N and M indicate the size of the board. Then goes N line, each line with M character shows the board. “#” Indicates the grass. You can assume that there is at least one grid which is consisting of grass in the board.
1 <= T <=100, 1 <= n <=10, 1 <= m <=10
Output
For each case, output the case number first, if they can play the MORE special (hentai) game (fire all the grass), output the minimal time they need to wait after they set fire, otherwise just output -1. See the sample input and output for more details.
Sample Input
Sample Output
Case 1:1
思路：枚举两起点BFS
#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <queue>using namespace std;#define ll long long#define INF 0x7fffffff#define MOD 1000007#define N 12struct node{ int x,y,t;};int tot;int n,m;int vis[N][N];char mpt[N][N];node start[N*N];queue<node> q;int dir[4][2]={1,0,-1,0,0,1,0,-1};int bfs(){ node now,next; while(!q.empty()) { now=q.front(); q.pop(); for(int i=0;i<4;i++) { next=now; next.t++; next.x+=dir[i][0]; next.y+=dir[i][1]; if(next.x>=1 && next.x<=n && next.y>=1 && next.y<=m && !vis[next.x][next.y] && mpt[next.x][next.y]==‘#‘) { q.push(next); vis[next.x][next.y]=1; } } } for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { if(mpt[i][j]==‘#‘ && !vis[i][j]) { return INF; } } } return now.t;}int main(){ int T,iCase=1; scanf("%d",&T); while(T--) { tot=0; scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) { scanf("%s",mpt[i]+1); } for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { if(mpt[i][j]==‘#‘) { start[++tot].x=i; start[tot].y=j; start[tot].t=0; } } } int ans=INF; for(int i=1;i<=tot;i++) { for(int j=i;j<=tot;j++) { while(!q.empty()) q.pop(); memset(vis,0,sizeof(vis)); q.push(start[i]); vis[start[i].x][start[i].y]=1; q.push(start[j]); vis[start[j].x][start[j].y]=1; ans=min(ans,bfs()); } } if(ans==INF) ans=-1; printf("Case %d: %d\n",iCase++,ans); } return 0;}
Description
Problem B: Fire!
Joe works in a maze. Unfortunately, portions of the maze have caught on fire, and the owner of the maze neglected to create a fire escape plan. Help Joe escape the maze.Given Joe‘s location in the maze and which squares of the maze are on fire, you must determine whether Joe can exit the maze before the fire reaches him, and how fast he can do it.
Joe and the fire each move one square per minute, vertically or horizontally (not diagonally). The fire spreads all four directions from each square that is on fire. Joe may exit the maze from any square that borders the edge of the maze. Neither Joe nor the fire may enter a square that is occupied by a wall.
Input Specification
The first line of input contains a single integer, the number of test cases to follow. The first line of each test case contains the two integers R and C, separated by spaces, with 1 <= R, C<= 1000. The following R lines of the test case each contain one row of the maze. Each of these lines contains exactly C characters, and each of these characters is one of:- #, a wall
- ., a passable square
- J, Joe‘s initial position in the maze, which is a passable square
- F, a square that is on fire
Sample Input
24 4#####JF##..##..#3 3####J.#.F
Output Specification
For each test case, output a single line containing IMPOSSIBLE if Joe cannot exit the maze before the fire reaches him, or an integer giving the earliest time Joe can safely exit the maze, in minutes.Output for Sample Input
3IMPOSSIBLE
思路：额有点坑、可能有多处火、加到队列BFS
#include <iostream>#include <cstdio>#include <cstring>#include <queue>using namespace std;#define INF 1<<30#define N 1010struct JF{ int x,y,t; JF(){} JF(int x,int y,int t):x(x),y(y),t(t){}};int n,m;int jx,jy;queue<JF> q;int vis[N][N];int fire[N][N]; //fire[i][j]表示火到i,j位置的最短时间char mpt[N][N];int dir[4][2]={1,0,-1,0,0,1,0,-1};bool judge(int x,int y){ if(x<1 || x>n || y<1 || y>m) return 0; if(mpt[x][y]==‘#‘) return 0; return 1;}void bfs1(){ JF now,next; while(!q.empty()) { now=q.front(); q.pop(); for(int i=0;i<4;i++) { next=now; next.t++; next.x+=dir[i][0]; next.y+=dir[i][1]; if(judge(next.x,next.y)) { if(next.t<fire[next.x][next.y]) { fire[next.x][next.y]=next.t; q.push(next); } } } }}void bfs2(){ queue<JF> q; memset(vis,0,sizeof(vis)); JF now,next; now.x=jx; now.y=jy; now.t=0; vis[jx][jy]=1; q.push(now); while(!q.empty()) { now=q.front(); q.pop(); if(now.x==1 || now.x==n || now.y==1 || now.y==m) { cout<<now.t+1<<endl; return; } for(int i=0;i<4;i++) { next=now; next.t++; next.x+=dir[i][0]; next.y+=dir[i][1]; if(judge(next.x,next.y) && !vis[next.x][next.y] && fire[next.x][next.y]-next.t>=1) { vis[next.x][next.y]=1; q.push(next); } } } cout<<"IMPOSSIBLE\n";}int main(){ int T; scanf("%d",&T); while(T--) { scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) { scanf("%s",mpt[i]+1); } for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { fire[i][j]=INF; if(mpt[i][j]==‘F‘) { fire[i][j]=0; q.push(JF(i,j,0)); } else if(mpt[i][j]==‘J‘) { jx=i; jy=j; } } } bfs1(); bfs2(); } return 0;}
Description
int maze[5][5] = {
0, 1, 0, 0, 0,
0, 1, 0, 1, 0,
0, 0, 0, 0, 0,
0, 1, 1, 1, 0,
0, 0, 0, 1, 0,
};
它表示一个迷宫，其中的1表示墙壁，0表示可以走的路，只能横着走或竖着走，不能斜着走，要求编程序找出从左上角到右下角的最短路线。
Input
Output
Sample Input
0 1 0 0 00 1 0 1 00 0 0 0 00 1 1 1 00 0 0 1 0
Sample Output
(0, 0)(1, 0)(2, 0)(2, 1)(2, 2)(2, 3)(2, 4)(3, 4)(4, 4)
思路：BFS+路径输出
#include <iostream>#include <cstdio>#include <queue>#include <cstring>using namespace std;#define mod 1000000007#define ll long long#define N 10struct node{ int x,y,t,pre;};struct queue{ node q[1010]; int front,rear;}q;int n=5;int vis[N][N];int mpt[N][N];int dir[4][2]={1,0,-1,0,0,1,0,-1};void show(int now){ int path[1010],len=0; path[++len]=now; while(q.q[now].pre!=-1) { path[++len]=q.q[now].pre; now=q.q[now].pre; } for(int i=len;i>=1;i--) { cout<<‘(‘<<q.q[path[i]].x-1<<", "<<q.q[path[i]].y-1<<‘)‘<<endl; }}void bfs(){ q.front=q.rear=0; memset(vis,0,sizeof(vis)); node now,next; now.x=1; now.y=1; now.t=0; now.pre=-1; vis[1][1]=1; q.q[++q.rear]=now; while(q.front!=q.rear) { now=q.q[++q.front]; if(now.x==n && now.y==n) { show(q.front); return; } for(int i=0;i<4;i++) { next=now; next.x+=dir[i][0]; next.y+=dir[i][1]; if(next.x>=1 && next.x<=n && next.y>=1 && next.y<=n && !vis[next.x][next.y] && !mpt[next.x][next.y]) { next.t++; next.pre=q.front; vis[next.x][next.y]=1; q.q[++q.rear]=next; } } }}int main(){ while(scanf("%d",&mpt[1][1])!=EOF) { for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { if(i==1 && j==1) continue; scanf("%d",&mpt[i][j]); } } bfs(); } return 0;}
Description
Input
Output
Sample Input
Sample Output
#include <iostream>#include <cstdio>#include <queue>#include <cstring>using namespace std;#define mod 1000000007#define ll long long#define N 110int n,m;char mpt[N][N];int dir[8][2]={1,1,-1,-1,1,-1,-1,1,1,0,-1,0,0,1,0,-1};void DFS(int x,int y){ for(int i=0;i<8;i++) { int tx=x+dir[i][0]; int ty=y+dir[i][1]; if(tx>=1 && ty>=1 && tx<=n && ty<=m && mpt[tx][ty]==‘@‘) { mpt[tx][ty]=‘*‘; DFS(tx,ty); } }}int main(){ int i,j; while(scanf("%d%d",&n,&m)!=EOF) { if(n==0 && m==0) break; for(i=1;i<=n;i++) { scanf("%s",mpt[i]+1); } int ans=0; for(i=1;i<=n;i++) { for(j=1;j<=m;j++) { if(mpt[i][j]==‘@‘) { ans++; mpt[i][j]=‘*‘; DFS(i,j); } } } cout<<ans<<endl; } return 0;}
Description
Input
Output
Sample Input
Sample Output
#include <iostream>#include <cstring>#include <queue>#include <cstdio>#include <algorithm>using namespace std;#define N 110struct node{ int w[3],t;};int v[3];int vis[N][N][N];void pour(node &t,int i,int j){ if(v[j]-t.w[j]>=t.w[i]) { t.w[j]+=t.w[i]; t.w[i]=0; } else { t.w[i]-=(v[j]-t.w[j]); t.w[j]=v[j]; }}bool judge(int a,int b,int c){ if(a==0 && b==c) return 1; if(b==0 && a==c) return 1; if(c==0 && b==a) return 1; return 0;}void bfs(){ node now,next; queue<node> q; memset(vis,0,sizeof(vis)); now.w[0]=v[0]; now.w[1]=0; now.w[2]=0; now.t=0; vis[v[0]][v[1]][v[2]]=1; q.push(now); while(!q.empty()) { now=q.front(); q.pop(); if(judge(now.w[0],now.w[1],now.w[2])) { cout<<now.t<<endl; return; } for(int i=0;i<3;i++) { for(int j=0;j<3;j++) { if(i==j) continue; next=now; pour(next,i,j); if(!vis[next.w[0]][next.w[1]][next.w[2]]) { next.t++; vis[next.w[0]][next.w[1]][next.w[2]]=1; q.push(next); } } } } cout<<"NO\n";}int main(){ while(scanf("%d%d%d",&v[0],&v[1],&v[2]),v[0]+v[1]+v[2]) { bfs(); } return 0;}
Description
Yifenfei’s home is at the countryside, but Merceki’s home is in the center of city. So yifenfei made arrangements with Merceki to meet at a KFC. There are many KFC in Ningbo, they want to choose one that let the total time to it be most smallest.
Now give you a Ningbo map, Both yifenfei and Merceki can move up, down ,left, right to the adjacent road by cost 11 minutes.
Input
Each test case include, first two integers n, m. (2<=n,m<=200).
Next n lines, each line included m character.
‘Y’ express yifenfei initial position.
‘M’ express Merceki initial position.
‘#’ forbid road;
‘.’ Road.
‘@’ KCF
Output
Sample Input
Sample Output
#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <queue>using namespace std;#define INF 0x7ffffff#define N 220struct node{ int x,y,t;};int n,m;int x[3],y[3];char mpt[N][N];int tmp[N][N][2];int vis[N][N];int dir[4][2]={1,0,-1,0,0,1,0,-1};void bfs(int index){ queue<node> q; memset(vis,0,sizeof(vis)); node now,next; now.x=x[index]; now.y=y[index]; now.t=0; vis[x[index]][y[index]]=1; tmp[x[index]][y[index]][index]=0; q.push(now); while(!q.empty()) { now=q.front(); q.pop(); for(int i=0;i<4;i++) { next=now; next.x+=dir[i][0]; next.y+=dir[i][1]; if(next.x>=1 && next.x<=n && next.y>=1 && next.y<=m && !vis[next.x][next.y] && mpt[next.x][next.y]!=‘#‘) { next.t++; if(mpt[next.x][next.y]==‘@‘) tmp[next.x][next.y][index]=next.t; vis[next.x][next.y]=1; q.push(next); } } }}int main(){ int i,j; while(scanf("%d%d",&n,&m)!=EOF) { for(i=1;i<=n;i++) scanf("%s",mpt[i]+1); for(i=1;i<=n;i++) { for(j=1;j<=m;j++) { tmp[i][j][1]=tmp[i][j][2]=INF; if(mpt[i][j]==‘M‘) x[1]=i,y[1]=j; else if(mpt[i][j]==‘Y‘) x[2]=i,y[2]=j; } } bfs(1); bfs(2); int ans=INF; for(i=1;i<=n;i++) { for(j=1;j<=m;j++) { if(mpt[i][j]==‘@‘) { ans=min(ans,tmp[i][j][1]+tmp[i][j][2]); } } } cout<<ans*11<<endl; } return 0;}
[kuangbin带你飞]专题一 简单搜索