走迷宫拿宝藏，拿到所有对应的钥匙才能开门
*解法：从起点bfs，遇到门时先放入队列中，取出的时候看钥匙够不够决定开不开门，如果不够就把它再放回队列继续往下走，当队列里只有几个门循环的时候就可以退出，所以记一个T<400
#include <iostream> #include <cstdio> #include <cstdlib> #include <queue> using namespace std; #define INF 1e9+10 char a[22][22]; int dx[] = {-1, 1, 0, 0}; int dy[] = {0, 0, -1, 1}; int vis[22][22], use[22][22]; int n, m, key[6], num[6], flag; queue<int> q; void bfs(int sx, int sy) { while(!q.empty()) q.pop(); vis[sx][sy] = 1; q.push(sx * m + sy); int T = 0; //队列里是能达到的点，标记vis的是能到达并拓展的点 while(!q.empty() && T < 400) { T++; int x = q.front() / m, y = q.front() % m; q.pop(); if(a[x][y] >= ‘A‘ && a[x][y] <= ‘E‘) { if(key[a[x][y] - ‘A‘] == num[a[x][y] - ‘A‘])//开门 { memset(vis, 0, sizeof(vis)); a[x][y] = ‘.‘; vis[x][y] = 1; } else//不开门 { q.push(x * m + y); continue; } } for(int i = 0; i < 4; i++) { int xx = x + dx[i], yy = y + dy[i]; if(xx >= 0 && xx < n && yy >= 0 && yy < m && !vis[xx][yy] && a[xx][yy] != ‘X‘) { if(a[xx][yy] == ‘G‘) {flag = 1; return;} if(a[xx][yy] == ‘.‘) { vis[xx][yy] = 1; q.push(xx * m + yy); } if(a[xx][yy] >= ‘a‘ && a[xx][yy] <= ‘e‘) { key[a[xx][yy] - ‘a‘]++; vis[xx][yy] = 1; a[xx][yy] = ‘.‘; q.push(xx * m + yy); } if(a[xx][yy] >= ‘A‘ && a[xx][yy] <= ‘E‘) { q.push(xx * m + yy);//放入队列但不标记 } } } } } int main() { while(1) { scanf("%d %d", &n, &m); if(n == 0 && m == 0) break; int sx, sy; memset(key, 0, sizeof(key)); memset(num, 0, sizeof(num)); memset(vis, 0, sizeof(vis)); memset(use, 0, sizeof(use)); for(int i = 0; i < n; i++) { scanf(" %s", a[i]); for(int j = 0; j < m; j++) { if(a[i][j] == ‘S‘) sx = i, sy = j; if(a[i][j] >= ‘a‘ && a[i][j] <= ‘e‘) num[a[i][j] - ‘a‘]++; } } for(int i = 0; i < 5; i++) if(num[i] == 0) num[i] = INF; flag = 0; bfs(sx, sy); if(flag) printf("YES\n"); else printf("NO\n"); } return 0; }