hdu2612——Find a way（BFS）-白红宇

hdu2612——Find a way（BFS）

阅读量：2345 次

发布时间：2019-05-10

本文共 1912 字，大约阅读时间需要 6 分钟。

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 4

Y.#@

….

.#..

@..M

4 4

Y.#@

….

.#..

@#.M

5 5

Y..@.

.#…

@..M.

#…#

Sample Output

两个人准备在某个KFC见面，由于时间是算两个人一共的和，所以要做两次BFS，计算每个人到某个KFC的最短时间，然后加起来比较最短的那个。

坑点是有的KFC可能有人到不了，所以要做好标记。。。

#include 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         #include 
          
           #define INF 0x3f3f3f3f#define MAXN 100010using namespace std;char mp[205][205];int dx[]= { 0,0,1,-1};int dy[]= { 1,-1,0,0};int n,m,visa[205][205],visb[205][205];struct Node{ int x,y,step;};map
           
            
             , int > h;void bfs(Node a,int vis[][205]){ vis[a.x][a.y]=1; queue
             
               q; q.push(a); while(!q.empty()) { Node tmp=q.front(); q.pop(); //cout<
              
               <<" "<
               
                <<" "<
                
                 <
                 
                  =0&&tmp1.x
                  
                   =0&&tmp1.y
                   
                    >mp[i][j]; if(mp[i][j]=='Y') { a.x=i; a.y=j; a.step=0; } if(mp[i][j]=='M') { b.x=i; b.y=j; b.step=0; } if(mp[i][j]=='@') { h[make_pair(i,j)]=0; } } bfs(a,visa); bfs(b,visb); int ans=INF; for(int i=0; i

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