M - Find a way

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博客讲述yifenfei回到宁波与好友Merceki相约KFC碰面，要选一个使两人总用时最短的KFC。给出宁波地图，两人可上下左右移动，每移动一格耗时11分钟。还给出输入输出格式及示例。

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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.
Input
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
Output
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.
Sample Input
4 4
Y.#@
…
.#…
@…M
4 4
Y.#@
…
.#…
@#.M
5 5
Y…@.
.#…
.#…
@…M.
#…#
Sample Output
66
88
66

#include<cstdio>
#include<cstring>
#include<queue>
#include<iostream>
#include<algorithm>
using namespace std;
typedef struct
{
	int x,y;
	int s;
}N;
N Y,M;
int n,m;
int vis[320][320];
char mise[320][320];
int res[320][320][5];
int d[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
void bfs(N t,int num)
{
	queue<N>q;
    q.push(t);
	memset(vis,0,sizeof(vis));
	vis[t.x][t.y]=1;
	N no,nn;
	while(q.size())
	{
      no=q.front();
	  q.pop();
	  if(mise[no.x][no.y]=='@')
	    res[no.x][no.y][num]=no.s;
	  for(int i=0;i<4;i++)
	  {
		  nn.x=no.x+d[i][0];
		  nn.y=no.y+d[i][1];
          nn.s=no.s+1;
		  if(nn.x>=0&&nn.x<n&&nn.y>=0&&nn.y<m&&!vis[nn.x][nn.y]&&mise[nn.x][nn.y]!='#')
		  {
			  vis[nn.x][nn.y]=1;
			   q.push(nn);
		  }
	  }
	}
}
int main()
{
	int i,j;
	while(cin>>n>>m)
	{ 
		for(i=0;i<n;i++)
		{
			 cin>>mise[i];
			 for(j=0;j<m;j++)
			 {
				 if(mise[i][j]=='Y')
				 {
					 Y.x=i;
					 Y.y=j;
					 Y.s=0;
				 }
				 if(mise[i][j]=='M')
				 {
					 M.x=i;
					 M.y=j;
                     M.s=0;
				 }
			 }
		 }
		 memset(res,0,sizeof(res));
		 bfs(Y,0);
		 bfs(M,1);
		 int ans=0x3f3f3f;
		 for(i=0;i<n;i++)
		 {
			 for(j=0;j<m;j++)
			 {
                if(res[i][j][0]&&res[i][j][1])
				ans=min(ans,res[i][j][0]+res[i][j][1]);
			 }
		 }
		 cout<<ans*11<<endl;
	}
	return 0;
}