HDU 2612 Find a way 简单广搜

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

Sample Input
4 4
Y.#@
....
.#..
@..M
4 4
Y.#@
....
.#..
@#.M
5 5
Y..@.
.#...
.#...
@..M.
#...#
        
 
Sample Output
66
88
66 

题意：

　　两个人想在某家KCF见面，要求他们到这家KCF的时间和花费最少。#表示不能通过；.表示可以通过；@表示KFC；Y、M分别表示两个人

思路：

　　分别以两个人为起点，走一遍广搜，然后将所有的数值全部记录下来，最后逐个便利取最小值。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#define MAX 0x3f3f3f3f
using namespace std;
struct point
{
    int x,y;
    int step;
} node;
int yx,yy,mx,my;
queue<point> q;
int vis[210][210];
char map[210][210];
int dir[4][2]= {-1,0, 1,0, 0,-1, 0,1};
int dis1[210][210],dis2[210][210];
int n,m;
void bfs(point node,int dis[][210])
{
    memset(vis,0,sizeof(vis));
    memset(dis,0,sizeof(dis));
    q.push(node);
    point now;
    while(!q.empty())
    {
        node=q.front();
        q.pop();
        for(int i=0; i<4; i++)
        {
            now.x=node.x+dir[i][0];
            now.y=node.y+dir[i][1];
            now.step=node.step+1;
            if(now.x>=0&&now.x<=m&&now.y>=0&&now.y<=n)
                if(map[now.x][now.y]!='#'&&vis[now.x][now.y]==0)
                {
                    vis[now.x][now.y]=1;
                    dis[now.x][now.y]=now.step;
                    q.push(now);
                }
        }
    }
}
int main()
{
    while(~scanf("%d %d",&m,&n))
    {
        getchar();
        for(int i=0; i<m; i++)
        {
            for(int j=0; j<n; j++)
            {
                scanf("%c",&map[i][j]);
                //cin>>map[i][j];
                if(map[i][j]=='Y')
                    yx=i,yy=j;
                if(map[i][j]=='M')
                    mx=i,my=j;
            }
            getchar();
        }
        //以第一个人为起点走一遍广搜，dis1记录第一个人到所有点的距离
        node.x=yx,node.y=yy,node.step=0;
        bfs(node,dis1);
        //以第二个人为起点走一遍广搜，dis2记录第二个人到所有点的距离
        node.x=mx,node.y=my,node.step=0;
        bfs(node,dis2);
        
        int min=MAX;
        for (int i=0; i<m; i++)
            for (int j=0; j<n; j++)
                if (map[i][j]=='@')
                    if (dis1[i][j]!=0&&dis2[i][j]!=0)
                        if(dis1[i][j]+dis2[i][j]<min)
                            min = dis1[i][j]+dis2[i][j];
        printf("%d
",min*11);
    }
}