[poj2762] Going from u to v or from v to u?(Kosaraju缩点+拓排)

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Going from u to v or from v to u?

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14778		Accepted: 3911

Description

In order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has n rooms, and one-way corridors connecting some rooms. Each time, Wind choose two rooms x and y, and ask one of their little sons go from one to the other. The son can either go from x to y, or from y to x. Wind promised that her tasks are all possible, but she actually doesn't know how to decide if a task is possible. To make her life easier, Jiajia decided to choose a cave in which every pair of rooms is a possible task. Given a cave, can you tell Jiajia whether Wind can randomly choose two rooms without worrying about anything?

Input

The first line contains a single integer T, the number of test cases. And followed T cases.

The first line for each case contains two integers n, m(0 < n
< 1001,m < 6000), the number of rooms and corridors in the cave.
The next m lines each contains two integers u and v, indicating that
there is a corridor connecting room u and room v directly.

Output

The output should contain T lines. Write 'Yes' if the cave has the property stated above, or 'No' otherwise.

Sample Input

1
3 3
1 2
2 3
3 1

Sample Output

Yes

题意：给一个图，问是否为单向连通。

Kosaraju+缩点，然后拓扑序搞一下，若只有一条没有分支的链，则Yes,否则No

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <vector>
 using namespace std;
 const int maxn=;
 vector<int>G[maxn];
 vector<int>rG[maxn];
 vector<int>vs;
 int vis[maxn];
 int cmp[maxn];
 int last[maxn];
 int deg[maxn];
 int next[maxn];
 int V;
 vector<int>vc[maxn];
 void add_edge(int u,int v)
 {
     G[u].push_back(v);
     rG[v].push_back(u);
 }
 void init(int n)
 {
     for(int i=;i<n;i++)
     {
         G[i].clear();
         rG[i].clear();
         vc[i].clear();
         vis[i]=;
         last[i]=-;
         deg[i]=;
         next[i]=-;
     }
 }
 void dfs(int v)
 {
     vis[v]=;
     for(int i=;i<G[v].size();i++)
     {
         if(!vis[G[v][i]])dfs(G[v][i]);
     }
     vs.push_back(v);
 }
 void rdfs(int v,int k)
 {
     vis[v]=;
     cmp[v]=k;
     vc[k].push_back(v);
     for(int i=;i<rG[v].size();i++)
     {
         if(!vis[rG[v][i]])rdfs(rG[v][i],k);
     }
 }
 int scc()
 {
     memset(vis,,sizeof(vis));
     vs.clear();
     for(int i=;i<V;i++)
     {
         if(!vis[i])dfs(i);
     }
     memset(vis,,sizeof(vis));
     int k=;
     for(int i=vs.size()-;i>=;i--)
     {
         if(!vis[vs[i]])rdfs(vs[i],k++);
     }
     return k;
 }
 int main()
 {
     ios::sync_with_stdio(false);
     int t;
     cin>>t;
     while(t--)
     {
         int n,m;
         cin>>n>>m;
         V=n;
         init(n);
         int u,v;
         for(int i=;i<m;i++)
         {
             cin>>u>>v;
             add_edge(--u,--v);
         }
         int k=scc();
         memset(vis,,sizeof(vis));
         int flag=;
         int num=;
         for(int i=;i<k;i++)
         {
             for(int j=;j<vc[i].size();j++)
             {
                 u=vc[i][j];
                 for(int l=;l<G[u].size();l++)
                 {
                     v=G[u][l];
                     if(cmp[u]!=cmp[v])
                     {
                         if(last[cmp[v]]==-)
                         {
                             last[cmp[v]]=cmp[u];
                         }
                         else if(last[cmp[v]]==cmp[u])continue;
                         else flag=;
                         if(next[cmp[u]]==-)
                         {
                             next[cmp[u]]=cmp[v];
                         }
                         else if(next[cmp[u]]==cmp[v])
                         {
                             continue;
                         }
                         else flag=;
                     }
                 }
             }
         }
         for(int i=;i<k;i++)
         {
             if(last[i]==-)num++;
         }
         if(flag&&num==)cout<<"Yes"<<endl;
         else cout<<"No"<<endl;
     }

     return ;
 }

代码君