Find a way (HDU 2612)

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

思路：
1.一开始使用从‘@‘跑到Y，M的方法，遍历每一个’@‘，即每一个’@‘都要跑一边bfs，然后就超时了
2.然后换成Y,M跑到’@‘，并分别记录下Y，M跑到每一个’@‘的时间，遍历’@'求出最小的和即可，这样只跑2遍

#include <iostream>
#include <queue>
#include <cstring>
#include <cstdio>
using namespace std;

int n,m;
char s[205][205];
int ans[3][205][205];
int X[4]={1,-1,0,0};
int Y[4]={0,0,1,-1};
bool visit[205][205];
struct node
{
    int x,y,step;
    node()
    {
        step=0;
    }
    node(int a,int b,int d)
    {
        x=a,y=b,step=d;
    }
}S,e[5],temp;


bool check(int x,int y)
{
    if(s[x][y]=='#'||x<1||x>n) return false;
    if(y>m||y<1||visit[x][y]) return false;
    return true;
}

void solve(int p)
{
    visit[e[p].x][e[p].y]=true;
    ans[p][e[p].x][e[p].y]=0;
    queue<node> que;
    que.push(e[p]);
    while(!que.empty())
    {
        temp=que.front();
        que.pop();
        for(int i=0;i<4;i++)
        {
            int nx=temp.x+X[i];
            int ny=temp.y+Y[i];
            if(check(nx,ny))
            {
                visit[nx][ny]=true;
                que.push({nx,ny,temp.step+1});
                if(s[nx][ny]=='@') ans[p][nx][ny]=temp.step+1;
            }
        }
    }
}

int main()
{
    std::ios::sync_with_stdio(false);
    cout.tie(0);
    while(cin>>n>>m)
    {
        memset(s,0,sizeof(s));
        memset(visit,0,sizeof(visit));
        memset(ans,0,sizeof(ans));
        int cnt=1;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                cin>>s[i][j];
                if(s[i][j]=='Y') e[1].x=i,e[1].y=j;
                if(s[i][j]=='M') e[2].x=i,e[2].y=j;
            }
        }
        int Ans=1000000;
        solve(1);
        memset(visit,0,sizeof(visit));
        solve(2);
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                if(s[i][j]=='@'&&ans[1][i][j]&&ans[2][i][j])
                    Ans=min(Ans,ans[1][i][j]+ans[2][i][j]);
            }
        }
        cout<<Ans*11<<'\n';
    }
    return 0;
}