hd 2612 Find a way （BFS）

Find a way

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10401    Accepted Submission(s): 3403

Problem 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 4Y.#@.....#..@..M4 4Y.#@.....#..@#.M5 5Y..@..#....#...@..M.#...#

 

Sample Output

668866
#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<queue>#define INF 0x3f3f3fusing namespace std;int dis1[210][210]; //用于储存Y到每个KFC的最小步数int dis2[210][210]; //用于储存M到每个KFC的最小步数char map[210][210];int vis[210][210];int n ,m;int dx[4] = {1,0,-1,0};int dy[4] = {0,1,0,-1};int sx1,sx2,sy1,sy2;struct node{     int x;     int y;     int step;};void bfs(int a,int b,int p){    memset(vis,0,sizeof(vis));    node now,next;    queue<node > q;    now.x = a;    now.y = b;    now.step = 0;    vis[now.x][now.y] = 1;    q.push(now);    while(!q.empty())    {        next = q.front();        q.pop();        if(map[next.x][next.y] == '@')        {            if(p == 1)                dis1[next.x][next.y] = next.step;            else                dis2[next.x][next.y] = next.step;        }        for(int i = 0 ; i < 4 ; i++)        {            now.x = next.x + dx[i];            now.y = next.y + dy[i];            if(now.x >= 0 && now.x < n && now.y >= 0 && now.y < m && map[now.x][now.y] != '#' && !vis[now.x][now.y])            {                vis[now.x][now.y] = 1;                now.step = next.step + 1;                q.push(now);            }        }    }}int main(){     while(cin >> n >> m)     {        int k = 0;        memset(vis,0,sizeof(vis));        memset(map,'\0',sizeof(map));        for(int i = 0 ; i < n ; i++)            cin >> map[i];        for(int i = 0 ; i < n ; i++)            for(int j = 0 ; j < m ; j++)           {               if(map[i][j] == 'Y')               {                   sx1 = i;                   sy1 = j;               }               if(map[i][j] == 'M')               {                   sx2 = i;                   sy2 = j;               }           }        memset(dis1,INF,sizeof(dis1)); // 每次计算都必须将dis1和dis2数组初始化        bfs(sx1,sy1,1);        memset(dis2,INF,sizeof(dis2));        bfs(sx2,sy2,2);        int MIN = INF;        for(int i = 0 ; i < n ; i++)            for(int j = 0 ; j < m ; j++)           {               if(dis1[i][j] != INF && dis2[i][j] != INF) //找到两个人都能到达的KFC                   MIN = min(dis1[i][j]+dis2[i][j],MIN);  //找出两人到达KFC的和的最小值           }        int temp = MIN * 11;        cout << temp << endl;    }    return 0;}