poj 2762强连通缩点加拓扑排序

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

先强连通缩点，再用拓扑排序，如果入度为0的点大于1个，就直接返回输出no，如果入度为0的点不超过2个输出yes 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#define maxn 1005
#define maxm 6005
using namespace std;
int low[maxn];
int dfn[maxn];
int head[maxn];
int sccno[maxn];
int dfsclock;
int sccnt;
int n,m,nm;
int in[maxn];
int cnt;
int t;
int u,v;
stack<int>s;
vector<int>g[maxn];
void init()
{
    memset(head,-1,sizeof(head));
    cnt=0;
}
struct edge
{
    int v,next;
}edges[maxm];
void addedge(int u,int v)
{
    edges[cnt].next=head[u];
    edges[cnt].v=v;
    head[u]=cnt++;
}
void dfs(int u)
{
    low[u]=dfn[u]=++dfsclock;
    s.push(u);
    for(int i=head[u];i!=-1;i=edges[i].next)
    {
        int v=edges[i].v;
        if(!dfn[v])
        {
            dfs(v);
            low[u]=min(low[u],low[v]);
        }
        else if(!sccno[v])
        {
            low[u]=min(low[u],dfn[v]);
        }
    }
    if(low[u]==dfn[u])
    {
        sccnt++;
        while(true)
        {
            int x=s.top();
            s.pop();
            sccno[x]=sccnt;
            if(x==u)
                break;
        }
    }
}
void findscc(int n)
{
    dfsclock=sccnt=0;
    memset(sccno,0,sizeof(sccno));
    memset(dfn,0,sizeof(dfn));
    for(int i=1;i<=n;i++)
        if(!dfn[i])
        dfs(i);
}
void top()
{
    queue<int>q;
    int sum=0;
    for(int i=1;i<=sccnt;i++)
        {if(in[i]==0)
        {sum++;
    q.push(i);
}
if(sum>1)
    {
        printf("No\n");
        return ;
    }
        }
while(!q.empty())
{
    int u=q.front();
    q.pop();
    sum=0;
    for(int i=0;i<g[u].size();i++)
    {
        int v=g[u][i];
        if(--in[v]==0)
        {
            sum++;

            if(sum>1)
                {
                    printf("No\n");
                    return;

                }
                q.push(v);
        }
    }
}

printf("Yes\n");
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {init();

        scanf("%d%d",&n,&m);
        while(m--)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            addedge(u,v);
        }
        findscc(n);
        for(int i=1;i<=sccnt;i++)
{g[i].clear();
in[i]=0;
}
        for(int u=1;u<=n;u++)
            for(int i=head[u];i!=-1;i=edges[i].next)
        {
            int v=edges[i].v;
        if(sccno[u]!=sccno[v])
        {
            in[sccno[v]]++;
            g[sccno[u]].push_back(sccno[v]);
        }
        }
       top();
    }
    return 0;
}