HDU2612 Find a way

解题思路：简单的宽搜题，分别存储Y，M到每个@的距离，然后求出每个@处

　　　　　对应累加的最小值。见代码：、

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
#define inf 0x3f3f3f3f
const int maxn = 205;
int vis[maxn][maxn], n, m, min1, dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};
int dis[maxn][maxn][2], flag;
char mapp[maxn][maxn];

struct node{
    int x, y;
    int cnt;
}s, e;

queue<node> q;

void bfs(int x, int y)
{
    memset(vis, 0, sizeof(vis)); //初始化
    s.x = x, s.y = y, s.cnt = 0;
    vis[x][y] = 1;
    while(!q.empty()) q.pop();
    q.push(s);

    while(!q.empty())
    {
        s = q.front(); q.pop();

        for(int i = 0; i < 4; i++)
        {
            e.x = s.x + dir[i][0];
            e.y = s.y + dir[i][1];
            e.cnt = s.cnt + 1;
            if(mapp[e.x][e.y] == '#' || vis[e.x][e.y]) continue;
            //存储下到每个@时对应的e.cnt的值
            if(mapp[e.x][e.y] == '@') dis[e.x][e.y][flag] = e.cnt;
           // dis[e.x][e.y][flag] = e.cnt;
            vis[e.x][e.y] = 1;
            q.push(e);
        }
    }
}

int main()
{
    while(~scanf("%d %d", &n, &m))
    {
        memset(mapp, '#', sizeof(mapp));
        memset(vis, 0, sizeof(vis));

        //刚开是这步初始化，结果WA了一发，好好思考为什么
        //memset(dis, 0, sizeof(dis));

        for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
        dis[i][j][0] = dis[i][j][1] = inf;

        for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++) scanf(" %c", &mapp[i][j]);

        min1 = inf;
        for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
        {
            if(mapp[i][j] == 'Y')
            {
                flag = 0; //将flag标记为0，表示Y到@
                bfs(i, j);
            }

            if(mapp[i][j] == 'M')
            {
                flag = 1;
                bfs(i, j);
            }
        }

        for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
        {
            //求出到每个@对应最小的时间
            if(mapp[i][j] == '@') min1 = min(min1, dis[i][j][0]+dis[i][j][1]);
        }
        printf("%d
", min1 * 11); //结果*11即可
    }
    return 0;
}