Super Phyllis（穷举+搜索）

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2723

题意：给出一些字符串u，v,代表u->v,问有几条边是多余的，也就是说去掉那些边后，u仍能到达v。

思路：穷举每条边，试着去掉该边，bfs搜索两个点是否仍然可达。

 #include<stdio.h>
 #include<iostream>
 #include<string.h>
 #include<map>
 #include<queue>
 #include<vector>
 #include<string>
 #include<algorithm>
 using namespace std;
 char s1[],s2[],s[][];
 map<string,int>f;
 vector<int>graph[];
 int Map[][],vis[];
 int n;
 int cnt;
 struct node
 {
     char s1[];
     char s2[];
 }edge[],tmp[];
 int cmp(const node a, const node b)
 {
     if(strcmp(a.s1,b.s1) == )
         return strcmp(a.s2,b.s2)<;
     return strcmp(a.s1,b.s1)<;
 }
 bool bfs(int s, int t)
 {
     queue<int> que;
     while(!que.empty())
         que.pop();
     vis[s] = ;
     que.push(s);
     while(!que.empty())
     {
         int u = que.front();
         que.pop();
         if(u == t)
             return true;
         for(int i = ; i < (int)graph[u].size(); i++)
         {
             int v = graph[u][i];
             if(!vis[v] && Map[u][v])
             {
                 que.push(v);
                 vis[v] = ;
             }
         }
     }
     return false;
 }

 int main()
 {
     int item = ;
     while(cin >> n)
     {
         if(n == ) break;
         for(int i = ; i <= ; i++)
             graph[i].clear();
         f.clear();
         memset(Map,,sizeof(Map));
         cnt = ;
         for(int i = ; i < n; i++)
         {
             cin>>s1>>s2;
             if(!f[s1])
             {
                 f[s1] = ++cnt;
                 strcpy(s[cnt],s1);
             }
             if(!f[s2])
             {
                 f[s2] = ++cnt;
                 strcpy(s[cnt],s2);
             }
             graph[ f[s1] ].push_back( f[s2] );
             Map[ f[s1] ][ f[s2] ] = ;
         }

         int res = ;
         for(int i = ; i <= cnt; i++)
         {
             for(int j = ; j < (int)graph[i].size(); j++)
             {
                 int v = graph[i][j];
                 memset(vis,,sizeof(vis));
                 Map[i][v] = ;
                 if(bfs(i,v))
                 {
                     strcpy(edge[res].s1,s[i]);
                     strcpy(edge[res].s2,s[v]);
                     res++;
                 }
                 else Map[i][v] = ;
             }
         }
         sort(edge,edge+res,cmp);
         printf("Case %d: ",item++);
         int t = ;
         if(res)
             tmp[t++] = edge[];
         for(int i = ; i < res; i++)
         {
             if(strcmp(edge[i].s1,edge[i-].s1) ==  && strcmp(edge[i].s2,edge[i-].s2) == )
                 continue;
             else
             {
                 tmp[t++] = edge[i];
             }
         }
         printf("%d",t);
         for(int i = ; i < t; i++)
         {
             printf(" %s,%s",tmp[i].s1,tmp[i].s2);
         }
         printf("\n");

     }
     return ;
 }