Find a way

Description
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
两个人初始位置是y和m，@代表咖啡馆，求哪个咖啡馆总用时最少。先搜y，然后找每个@的ans1时间，首先他还得能访问到，所以加一个leap数组，然后就是ans2，重复操作，记得初始化，最后找最小值就好了。

#include
#include
#include
#include
#include
using namespace std;
char a[205][205];
int leap[205][205], m, n, x1, yo, x2, y2, num, time;
int dir[4][2] = { 1, 0, -1, 0, 0, 1, 0, -1 };
int dp[205][205];
struct Node{
    int x;
    int y;
    int ans1;
    int ans2;
    int leap;
}node[4050];
queuep;
int min(int a, int b)
{
    return a < b ? a : b;
}
void bfs(int x1,int y1)
{
    Node temp, now;
    int i, tempx, tempy;
    temp.x = x1; temp.y = y1; p.push(temp);
    memset(leap, 0, sizeof(leap));
    leap[x1][y1] = 1;
    dp[x1][y1] = 0;
    while (p.size())
    {
        temp = p.front();
        for (i = 0; i < 4; i++)
        {
            tempx = temp.x + dir[i][0];
            tempy = temp.y + dir[i][1];
            if (tempx >= 1 && tempx <= n&&tempy >= 1 && tempy <= m)
            {
                if (!leap[tempx][tempy] && a[tempx][tempy] != '#')
                {
                    leap[tempx][tempy] = 1;
                    now.x = tempx; now.y = tempy;
                    dp[tempx][tempy] = dp[temp.x][temp.y] + 1;
                    p.push(now);
                }
            }
        }
        p.pop();
    }
}
int main()
{
    int i, j, ans, k, time1, time2;
    while (scanf("%d%d", &n, &m) != EOF)
    {
        getchar(); k = 0;
        for (i = 1; i <= n; i++)
        {
            for (j = 1; j <= m; j++)
            {
                cin >> a[i][j];
                if (a[i][j] == 'Y')
                {
                    x1 = i; yo = j;
                }
                if (a[i][j] == 'M')
                {
                    x2 = i; y2 = j;
                }
                if (a[i][j] == '@')
                {
                    k++;
                    node[k].x = i; node[k].y = j; node[k].leap = 1;
                }
            }
            getchar();
        }
        num = k;
        time = 999;
        bfs(x1, yo);
        for (i = 1; i <= num; i++)
        if (leap[node[i].x][node[i].y])
            node[i].ans1 = dp[node[i].x][node[i].y];
        else
            node[i].leap = 0;
        bfs(x2, y2);
        for (i = 1; i<= num; i++)
        if (leap[node[i].x][node[i].y])
            node[i].ans2 = dp[node[i].x][node[i].y];
        else
            node[i].leap = 0;
        for (i = 1; i <= num; i++)
        {
            if (node[i].leap == 0)continue;
            time = min(time, node[i].ans1 + node[i].ans2);
        }
        cout << time*11 << endl;
    }
    return 0;
}