题目连接：Find a way

 题目大意：Y 和 M 想在 KFC 见面，找到一条最短的路，其中 # 是障碍，不能走这里。

思路：保存每一个 KFC 的位置，计算从每一个 KFC 到这两个人的距离即可。。。

But ... 为啥会 MLE ！

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
struct Node{
	int x;
	int y;
	int step;
};
int n,m,ans,minn;
char a[201][201];
int vis[201][201];
int xx[4] = {0,0,1,-1};
int yy[4] = {1,-1,0,0};
queue<Node>q,que;
int pd(int x,int y)
{
	if(x >= 1 && x <= n && y >= 1 && y <= m && a[x][y] != '#')
	return 1;
	return 0;
}
int main()
{
	while(scanf("%d%d",&n,&m) != EOF)
	{
		int tot = 0;
		while(!q.empty())
			q.pop();
		Node st,en;
		for(int i=1;i<=n;i++)
		{
			for(int j=1;j<=m;j++)
			{
				cin>>a[i][j];
				if(a[i][j] == '@')
				{
					Node now;
					now.x = i;
					now.y = j;
					q.push(now);
					a[i][j] = '#';
				}
				if(a[i][j] == 'Y')
				{
					st.x = i;
					st.y = j;
					a[i][j] == '.';
				}
				if(a[i][j] == 'M')
				{
					en.x = i;
					en.y = j;
					a[i][j] == '.';
				}
			}
		}
		minn = 99999999;
		while(!q.empty())
		{
			while(!que.empty())
				que.pop();
			memset(vis,0,sizeof(vis));
			Node no = q.front();
			q.pop();
			no.step = 0;
			que.push(no);
			int sum = 0;
			bool flag1 = 0,flag2 = 0;
			ans = 0;
			while(!que.empty())
			{
				Node now = que.front();
				que.pop();
				vis[now.x][now.y] = 1;
				if(now.x == st.x && now.y == st.y && !flag1)
				{
					flag1 = 1;
					ans += now.step;
					//cout<<"First: "<<now.step<<endl;
				}
				if(now.x == en.x && now.y == en.y && !flag2)
				{
					flag2 = 1;
					ans += now.step;
					//cout<<"Secend: "<<now.step<<endl;
				}
				if(flag1 && flag2){
					//cout<<"minn = "<<minn<<endl;
					//cout<<"ans = "<<ans<<endl<<endl; 
					if(minn > ans)
					minn = ans;
					break;
				}
				for(int i=0;i<4;i++)
				{
					int tx = now.x + xx[i];
					int ty = now.y + yy[i];
					if(pd(tx,ty) && !vis[tx][ty])
					{
						Node nex;
						nex.x = tx;
						nex.y = ty;
						nex.step = now.step + 1;
						que.push(nex);
					}
				}
			}
		}
		cout<<minn*11<<endl;
	}
	return 0;
}