Find a way

Pass a year learning in Hangzhou, yifenfei arrival hometown Ningbo at finally. Leave Ningbo one year, yifenfei have many people to meet. Especially a good friend Merceki. 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.

The input contains multiple test cases. 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

For each test case output the minimum total time that both yifenfei and Merceki to arrival one of KFC.You may sure there is always have a KFC that can let them meet.

4 4Y.#@.....#..@..M4 4Y.#@.....#..@#.M5 5Y..@..#....#...@..M.#...#

668866  
    
    
    
     
   /*分析：直接想到两次bfs然后扫一次求点最小 有个重点的地方！就是那点必须是两个人都能走到的！！wa了2次 */#include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         using namespace std;const int N=200+5;char map[N][N];bool book[N][N];short ok[N][N];int re[N][N];int n,m;struct Node{	int x,y,step;}node,tmp;queue
         
           q;int step[4][2]={ {1,0},{0,1},{-1,0},{0,-1}};inline void bfs(int bx,int by){ memset(book,0,sizeof(book)); while(!q.empty()) q.pop(); node.x=bx; node.y=by; node.step=0; q.push(node); book[bx][by]=1; while(!q.empty()){ node=q.front(); q.pop(); if(map[node.x][node.y]=='@'){ ok[node.x][node.y]++; re[node.x][node.y]+=node.step; } int i; for(i=0;i<4;i++){ int a=node.x+step[i][0]; int b=node.y+step[i][1]; if(a<0||a>=n||b<0||b>=m||book[a][b]||map[a][b]=='#' /* ||map[a][b]=='M'||map[a][b]=='Y'*/) continue; book[a][b]=1; tmp.x=a; tmp.y=b; tmp.step=node.step+11; q.push(tmp); } }}int main(){ int i,j,x1,y1,x2,y2; while(scanf("%d%d",&n,&m)==2){ memset(re,0,sizeof(re)); memset(ok,0,sizeof(ok)); for(i=0;i
          
           re[i][j]) small=re[i][j]; printf("%d\n",small); } return 0;}