解题思路：简单的宽搜题，分别存储Y，M到每个@的距离，然后求出每个@处
对应累加的最小值。见代码：、
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<queue> 5 using namespace std; 6 #define inf 0x3f3f3f3f 7 const int maxn = 205; 8 int vis[maxn][maxn], n, m, min1, dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1}; 9 int dis[maxn][maxn][2], flag; 10 char mapp[maxn][maxn]; 11 12 struct node{ 13 int x, y; 14 int cnt; 15 }s, e; 16 17 queue<node> q; 18 19 void bfs(int x, int y) 20 { 21 memset(vis, 0, sizeof(vis)); //初始化 22 s.x = x, s.y = y, s.cnt = 0; 23 vis[x][y] = 1; 24 while(!q.empty()) q.pop(); 25 q.push(s); 26 27 while(!q.empty()) 28 { 29 s = q.front(); q.pop(); 30 31 for(int i = 0; i < 4; i++) 32 { 33 e.x = s.x + dir[i][0]; 34 e.y = s.y + dir[i][1]; 35 e.cnt = s.cnt + 1; 36 if(mapp[e.x][e.y] == '#' || vis[e.x][e.y]) continue; 37 //存储下到每个@时对应的e.cnt的值 38 if(mapp[e.x][e.y] == '@') dis[e.x][e.y][flag] = e.cnt; 39 // dis[e.x][e.y][flag] = e.cnt; 40 vis[e.x][e.y] = 1; 41 q.push(e); 42 } 43 } 44 } 45 46 int main() 47 { 48 while(~scanf("%d %d", &n, &m)) 49 { 50 memset(mapp, '#', sizeof(mapp)); 51 memset(vis, 0, sizeof(vis)); 52 53 //刚开是这步初始化，结果WA了一发，好好思考为什么 54 //memset(dis, 0, sizeof(dis)); 55 56 for(int i = 1; i <= n; i++) 57 for(int j = 1; j <= m; j++) 58 dis[i][j][0] = dis[i][j][1] = inf; 59 60 for(int i = 1; i <= n; i++) 61 for(int j = 1; j <= m; j++) scanf(" %c", &mapp[i][j]); 62 63 min1 = inf; 64 for(int i = 1; i <= n; i++) 65 for(int j = 1; j <= m; j++) 66 { 67 if(mapp[i][j] == 'Y') 68 { 69 flag = 0; //将flag标记为0，表示Y到@ 70 bfs(i, j); 71 } 72 73 if(mapp[i][j] == 'M') 74 { 75 flag = 1; 76 bfs(i, j); 77 } 78 } 79 80 for(int i = 1; i <= n; i++) 81 for(int j = 1; j <= m; j++) 82 { 83 //求出到每个@对应最小的时间 84 if(mapp[i][j] == '@') min1 = min(min1, dis[i][j][0]+dis[i][j][1]); 85 } 86 printf("%d ", min1 * 11); //结果*11即可 87 } 88 return 0; 89 }