题解：
M和Y约在KFC见面，问两个人去到同一家KFC所用的最短时间，直接求出M和Y到所有点的最短时间，然后枚举到KFC的最小值即可
代码
#include <cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <vector>
#include <map>
#include <string>
#include <set>
#include <ctime>
#include <cmath>
#include <cctype>
using namespace std;
#define MAX 100000
#define LL long long
int cas=1,T;
char mapp[201][201];
int vis[201][201];
int county[201][201];
int countm[201][201];
struct Node
{
int x;
int y;
int countt;
};
queue<Node> que;
int n,m;
bool check(int x,int y)
{
if (x<0 || y<0 || x>n || y>m)
return true;
if (vis[x][y] || mapp[x][y]=='#')
return true;
return false;
}
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
int main()
{
while (scanf("%d%d",&n,&m)!=EOF) //忘记加 !=EOF TLE两次！
{
Node ys,ms;
for (int i = 0;i<n;i++)
scanf("%s",&mapp[i]);
for (int i = 0;i<n;i++)
for (int j = 0;j<m;j++)
{
if (mapp[i][j]=='Y')
{
ys.x=i;
ys.y=j;
ys.countt=0;
}
if (mapp[i][j]=='M')
{
ms.x=i;
ms.y=j;
ms.countt=0;
}
}
vis[ys.x][ys.y]=1;
que.push(ys);
while (!que.empty())
{
Node t=que.front();
que.pop();
for (int i = 0;i<4;i++)
{
Node temp;
temp.x = t.x+dir[i][0];
temp.y = t.y+dir[i][1];
temp.countt=t.countt+1;
if (check(temp.x,temp.y))
continue;
vis[temp.x][temp.y]=1;
county[temp.x][temp.y]=temp.countt;
que.push(temp);
}
}
while (!que.empty())
que.pop();
memset(vis,0,sizeof(vis));
vis[ms.x][ms.y]=1;
que.push(ms);
while (!que.empty())
{
Node t = que.front();
que.pop();
for (int i = 0;i<4;i++)
{
Node temp;
temp.x = t.x+dir[i][0];
temp.y = t.y+dir[i][1];
temp.countt=t.countt+1;
if (check(temp.x,temp.y))
continue;
vis[temp.x][temp.y]=1;
countm[temp.x][temp.y]=temp.countt;
que.push(temp);
}
}
int mins = 99999999;
for (int i = 0;i<n;i++)
for (int j = 0;j<m;j++)
{
if (mapp[i][j]=='@')
if (county[i][j]+countm[i][j] < mins)
mins=county[i][j] + countm[i][j];
}
printf("%d\n",mins*11);
memset(mapp,0,sizeof(mapp));
memset(countm,0,sizeof(countm));
memset(county,0,sizeof(county));
memset(vis,0,sizeof(vis));
while (!que.empty())
que.pop();
}
return 0;
}
