1.題意：你約你喜歡的妹子去KFC，你們想找一家最近的，使得你和妹子到KFC所用時間和最短。

2.思路：相當於進行兩次BFS搜尋，可以從每個KFC當做起始點 出發搜尋Y和M，也可以從Y和M出發搜尋到達每個KFC的時間，然後用一個數組儲存到達每個kfc的總時間，找到最小值輸出即可，要記得在合適的時間初始化vis和cost陣列，要不然會出現不必要的麻煩。要注意考慮到題目中的極端情況，也就是有一家KFC距離最近，但是無法到達，所以不能夠選擇這一家，考慮到這一點之後還是能比較輕鬆的AC的

3.貼上程式碼：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
#include <queue>
#include <string>
using namespace std;
char Map[205][205];
int vis[205][205],cost[205][205];
int mov1[4]{1,-1,0, 0};
int mov2[4]{0, 0,1,-1};
int row,col,yx,yy,mx,my,ans;
struct node{
    int x,y,step;
};
int bfs(int a,int b)
{
    memset(vis,0,sizeof(vis));
    queue<node> way;
    node start;
    start.x=a;
    start.y=b;
    start.step=0;
    way.push(start);
    while(!way.empty())
    {
        node now=way.front();
        way.pop();
        int xx=now.x,yy=now.y,ss=now.step;
        //cout<<xx<<"  "<<yy<<" "<<ss<<endl;
        for(int i=0;i<4;i++)
        {
            if(vis[xx+mov1[i]][yy+mov2[i]]!=1&&
               xx+mov1[i]>=0&&xx+mov1[i]<row
               &&yy+mov2[i]>=0&&yy+mov2[i]<col)
            {
                if(Map[xx+mov1[i]][yy+mov2[i]]=='#')
                    continue;
                if(Map[xx+mov1[i]][yy+mov2[i]]=='@')
                {
                    node newnode;
                    vis[xx+mov1[i]][yy+mov2[i]]=1;
                    newnode.x=xx+mov1[i];
                    newnode.y=yy+mov2[i];
                    newnode.step=ss+1;
                    way.push(newnode);
                    cost[xx+mov1[i]][yy+mov2[i]]+=newnode.step;
                    //cout<<cost[newnode.x][newnode.y]<<endl;
                }
                else if(Map[xx+mov1[i]][yy+mov2[i]]=='.')
                {
                    node newnode;
                    vis[xx+mov1[i]][yy+mov2[i]]=1;
                    newnode.x=xx+mov1[i];
                    newnode.y=yy+mov2[i];
                    newnode.step=ss+1;
                    way.push(newnode);
                }
            }
        }
    }
    int mmax=1000000;
    for(int i=0;i<row;i++){
        for(int j=0;j<col;j++){
            if(cost[i][j]!=0&&cost[i][j]<mmax)
                mmax=cost[i][j];
            //cout<<cost[i][j]<<" ";
        }
        //cout<<endl;
    }
    //cout<<mmax<<endl;
    return mmax;
}
int main()
{
    string s;
    while(~scanf("%d%d",&row,&col))
    {
        memset(cost,0,sizeof(cost));
        ans=0;
        for(int i=0;i<row;i++)
        {
            cin>>s;
            for(int j=0;j<col;j++)
            {
                Map[i][j]=s[j];
                if(s[j]=='Y'){
                    yx=i;yy=j;
                }
                else if(s[j]=='M'){
                    mx=i;my=j;
                }
            }
        }
        ans=bfs(yx,yy);
        //cout<<ans<<"######"<<endl;
        ans=bfs(mx,my);
        //cout<<ans<<"####33"<<endl;
        printf("%d\n",ans*11);
    }
    return 0;
}