[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=1010;
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=0;i<n;i++)
    {
        G[i].clear();
        rG[i].clear();
        vc[i].clear();
        vis[i]=0;
        last[i]=-1;
        deg[i]=0;
        next[i]=-1;
    }
}
void dfs(int v)
{
    vis[v]=1;
    for(int i=0;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]=1;
    cmp[v]=k;
    vc[k].push_back(v);
    for(int i=0;i<rG[v].size();i++)
    {
        if(!vis[rG[v][i]])rdfs(rG[v][i],k);
    }
}
int scc()
{
    memset(vis,0,sizeof(vis));
    vs.clear();
    for(int i=0;i<V;i++)
    {
        if(!vis[i])dfs(i);
    }
    memset(vis,0,sizeof(vis));
    int k=0;
    for(int i=vs.size()-1;i>=0;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=0;i<m;i++)
        {
            cin>>u>>v;
            add_edge(--u,--v);
        }
        int k=scc();
        memset(vis,0,sizeof(vis));
        int flag=1;
        int num=0;
        for(int i=0;i<k;i++)
        {
            for(int j=0;j<vc[i].size();j++)
            {
                u=vc[i][j];
                for(int l=0;l<G[u].size();l++)
                {
                    v=G[u][l];
                    if(cmp[u]!=cmp[v])
                    {
                        if(last[cmp[v]]==-1)
                        {
                            last[cmp[v]]=cmp[u];
                        }
                        else if(last[cmp[v]]==cmp[u])continue;
                        else flag=0;
                        if(next[cmp[u]]==-1)
                        {
                            next[cmp[u]]=cmp[v];
                        }
                        else if(next[cmp[u]]==cmp[v])
                        {
                            continue;
                        }
                        else flag=0;
                    }
                }
            }
        }
        for(int i=0;i<k;i++)
        {
            if(last[i]==-1)num++;
        }
        if(flag&&num==1)cout<<"Yes"<<endl;
        else cout<<"No"<<endl;
    }

    return 0;
}