hdu 4460(STL+BFS)-白红宇

hdu 4460(STL+BFS)

阅读量：5313 次

发布时间：2019-06-14

本文共 1153 字，大约阅读时间需要 3 分钟。

Problem analysis:

For this problem,first give you some people's name and then give you the relations of them.

Now,you task is calculate the maximum steps between the every two guys.

To solve the problem,you just need to calculate all the steps between the every two guys.

And of course,the BFS is good to use.

Here is the detailed code:

#include
     
      #include
      
       #include
       
        #include
        
         #include
         using namespace std;const int M=1020;const int INF=0x3f3f3f3f3f;int dis[M][M];//记录每两个人之间的距离int mark[M];int n,m;map
          
           match;vector
           
            vec[M];queue
            
             q;void BFS(int x)//实现标号为x的人的朋友关系搜索{ memset(mark,0,sizeof(mark)); dis[x][x]=0; mark[x]=1; q.push(x); while(!q.empty()) { int t=q.front(); q.pop(); int y=vec[t].size();//统计标号为t的人的直接朋友人数 for(int j=0;j
             
              >n) { if(n==0) break; int i,j; for(i=0;i
              
               >name; match[name]=i; } for(i=0;i
               
                >m; for(i=0;i
                
                 >name1>>name2; int t1=match[name1]; int t2=match[name2]; vec[t1].push_back(t2); vec[t2].push_back(t1); } for(i=0;i
                 
                  ans) ans=dis[i][j]; } } if(ans==INF) ans=-1; cout<
                  
                   <