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HDU 2102 A计划 (BFS或DFS)

2024-10-22 22:36:39   209人阅读

题意：中文题。

析：是一个简单的搜索，BFS 和 DFS都可行， 主要是这个题有一个坑点，那就是如果有一层是#，另一个层是#或者*，都是过不去的，就可以直接路过，

剩下的就是一个简单的搜索，只不过是两层而已，可能一个就是在#必须传送，这个题目已经说的很清楚了。

代码如下：

BFS：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 10 + 10;
const int mod = 1000000007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}

char s[2][maxn][maxn];
int vis[2][maxn][maxn];
int t;
struct Node{
  int pos, x, y;
  Node() { }
  Node(int p, int xx, int yy) : pos(p), x(xx), y(yy) { }
  bool operator == (const Node &p) const{
    return pos == p.pos && x == p.x && y == p.y;
  }
};

bool bfs(){
  memset(vis, -1, sizeof vis);
  Node goal;
  for(int i = 0; i < n; ++i)
    for(int j = 0; j < m; ++j)
      if(s[0][i][j] == ‘P‘)  goal = Node(0, i, j);
      else if(s[1][i][j] == ‘P‘)  goal = Node(1, i, j);
  queue<Node> q;
  vis[0][0][0] = 0;
  q.push(Node(0, 0, 0));

  while(!q.empty()){
    Node u = q.front();  q.pop();
    if(vis[u.pos][u.x][u.y] > t)  return false;
    if(u == goal)  return true;
    for(int i = 0; i < 4; ++i){
      int x = u.x + dr[i];
      int y = u.y + dc[i];
      if(!is_in(x, y) || vis[u.pos][x][y] != -1 || s[u.pos][x][y] == ‘*‘)  continue;
      vis[u.pos][x][y] = vis[u.pos][u.x][u.y] + 1;
      if(s[u.pos][x][y] == ‘#‘){
        vis[u.pos^1][x][y] = vis[u.pos][x][y];
        if(s[u.pos^1][x][y] == ‘#‘ || s[u.pos^1][x][y] == ‘*‘)  continue;
        q.push(Node(u.pos^1, x, y));
        continue;
      }
      q.push(Node(u.pos, x, y));
    }
  }
  return false;
}

int main(){
  int T;  cin >> T;
  while(T--){
    scanf("%d %d %d", &n, &m, &t);
    for(int i = 0; i < n; ++i)  scanf("%s", s[0]+i);
    for(int i = 0; i < n; ++i)  scanf("%s", s[1]+i);
    bool ans = bfs();
    printf("%s\n", ans ? "YES" : "NO");
  }
  return 0;
}

 

　DFS：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 10 + 10;
const int mod = 1000000007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}

char s[2][maxn][maxn];
bool vis[2][maxn][maxn];
int t;

bool dfs(int pos, int r, int c, int d){
  if(d > t)  return false;
  if(s[pos][r][c] == ‘P‘)  return true;
  for(int i = 0; i < 4; ++i){
    int x = r + dr[i];
    int y = c + dc[i];
    if(!is_in(x, y) || vis[pos][x][y] || s[pos][x][y] == ‘*‘)  continue;
    vis[pos][x][y] = 1;
    if(s[pos][x][y] == ‘#‘){
      vis[pos^1][x][y] = 1;
      if(dfs(pos^1, x, y, d+1))  return true;
      vis[pos^1][x][y] = 0;
    }
    else if(dfs(pos, x, y, d+1))  return true;
    vis[pos][x][y] = 0;
  }
  return false;
}


int main(){
  int T;  cin >> T;
  while(T--){
    scanf("%d %d %d", &n, &m, &t);
    for(int i = 0; i < n; ++i)  scanf("%s", s[0]+i);
    for(int i = 0; i < n; ++i)  scanf("%s", s[1]+i);
    for(int i = 0; i < n; ++i)
      for(int j = 0; j < m; ++j)
        if(s[0][i][j] == ‘#‘ && (s[1][i][j] == ‘*‘ || s[1][i][j] == ‘#‘))  s[0][i][j] = s[1][i][j] = ‘*‘;
        else if(s[0][i][j] == ‘*‘ && s[1][i][j] == ‘#‘)  s[1][i][j] = ‘*‘;
    memset(vis, 0, sizeof vis);
    vis[0][0][0] = 1;
    bool ans = dfs(0, 0, 0, 0);
    printf("%s\n", ans ? "YES" : "NO");
  }
  return 0;
}

　　

 

HDU 2102 A计划 (BFS或DFS)


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