Hdu 2612-Find a way(bfs)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2612

题目大意：就是Y和M要到一个KFC商谈， 算出Y和M到KFC花费的时间和最小

题目思路：直接bfs

代码如下：

#include<bits/stdc++.h>
using namespace std;

struct node
{
    int x, y, t;
    node(int a, int b, int c)
    {
        x = a, y = b, t = c;
    }
};

char mp[203][203];
int arr[203][203], may[203][203];
bool mk[203][203];
int dir[4][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };

void bfs(int bgx, int bgy)
{
    memset(mk, false, sizeof(mk));
    queue<node> que;
    que.push(node(bgx, bgy, 0));
    mk[bgx][bgy] = true;
    while( !que.empty())
    {
        node f = que.front();
        que.pop();
        for(int i=0; i<4; ++ i)
        {
            int x = f.x + dir[i][0], y = f.y + dir[i][1];
            int t = f.t + 1;
            if(mp[x][y] != '#' && mk[x][y] == false)
            {
                mk[x][y] = true;
                if(mp[x][y] == '@')
                    arr[x][y] += t, may[x][y] += 1;
                que.push(node(x, y, t));
            }
        }
    }
}
int main()
{
    int n, m;
    while(scanf("%d%d", &n, &m) != EOF)
    {
        int y_bgx, y_bgy, m_bgx, m_bgy;
        memset(mp, '#', sizeof(mp));
        for(int i=1; i<=n; ++ i)
        {
            for(int j=1; j<=m; ++ j)
            {
                scanf(" %c", &mp[i][j]);
                if(mp[i][j] == 'Y')
                    y_bgx = i, y_bgy = j;
                if(mp[i][j] == 'M')
                    m_bgx = i, m_bgy = j;
            }
        }
        memset(arr, 0, sizeof(arr));
        memset(may, 0, sizeof(may));

        bfs(y_bgx, y_bgy);
        bfs(m_bgx, m_bgy);
        int ans = 1000000;
        for(int i=1; i<=n; ++ i)
        {
            for(int j=1; j<=m; ++ j)
            {
                if(mp[i][j] == '@' && may[i][j] == 2)
                {
                    if(arr[i][j] < ans)
                        ans = arr[i][j];
                }
            }
        }
        printf("%d\n", ans * 11);
    }
    return 0;
}