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题目分析

        1.双重bfs,遍历两个起点求最短路再计算总和即可

        2.唯一的坑点在于对于一个KFC，两人中可能有一个到不了，所以还要对到不了的点距离做处理

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#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 220;

struct pos{
    int y, x;
}Y, M;

char g[N][N];
bool vis[N][N];
int disy[N][N];
int dism[N][N];
int t1, t2;

int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};

void bfs1()
{
    memset(vis, 0, sizeof vis);
    queue<pos> q;
    q.push(Y);
    vis[Y.y][Y.x] = 1;
    while(!q.empty())
    {
        pos temp = q.front(); q.pop();
        for(int i = 0; i < 4; i++)
        {
            int a = temp.x + dx[i]; int b = temp.y + dy[i];
            if(a < 1 || b < 1 || a > t2 || b > t1) continue;
            if(!vis[b][a] && g[b][a] != '#')
            {
                vis[b][a] = 1;
                q.push({b, a});
                disy[b][a] = disy[temp.y][temp.x] + 1;
            }
        }
    }
        
                for(int i = 1; i <= t1; i++)
                {
                    for(int j = 1; j <= t2; j++)
                    {
                        if(disy[i][j] == 0) disy[i][j] = 1e7;
                    }
                }
}

void bfs2()
{
    memset(vis, 0, sizeof vis);
    queue<pos> q;
    q.push(M);
    vis[M.y][M.x] = 1;
    
    while(!q.empty())
    {
        pos temp = q.front(); q.pop();
        for(int i = 0; i < 4; i++)
        {
            int a = temp.x + dx[i]; int b = temp.y + dy[i];
            if(a < 1 || b < 1 || a > t2 || b > t1) continue;
            if(!vis[b][a] && g[b][a] != '#')
            {
                vis[b][a] = 1;
                q.push({b, a});
                dism[b][a] = dism[temp.y][temp.x] + 1;
            }
        }
    }
    
        for(int i = 1; i <= t1; i++)
        {
            for(int j = 1; j <= t2; j++)
            {
                if(dism[i][j] == 0) dism[i][j] = 1e7;
            }
        }
        
}

int main()
{
    while(scanf("%d %d", &t1, &t2) != EOF)
    {
        memset(disy, 0, sizeof disy);
        memset(dism, 0, sizeof dism);
        for(int i = 1; i <= t1; i++) 
            for(int j = 1; j <= t2; j++)
            {
                scanf(" %c", &g[i][j]);
                if(g[i][j] == 'Y') Y.x = j, Y.y = i;
                else if(g[i][j] == 'M') M.x = j, M.y = i;
            }
        
        bfs1();
        bfs2();
        
       int ans = 999;
        for(int i= 1; i <= t1; i++)
        {
            for(int j= 1; j <= t2; j++)
            {
                if(g[i][j] == '@') ans = min(ans, disy[i][j] + dism[i][j]);
            }
        }
        printf("%d\n", ans * 11);
    }

}