【图论补完计划】poj 2762 （强连通分量 kosaraju）

Going from u to v or from v to u?

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 17479		Accepted: 4694

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
询问一个有向图的任意两点s，t，是否存在s可达t或t可达s，如果都存在输出Yes，否则输出No。

做法
scc拆点，建图，记录入度，如果存在两个及两个以上的入度为0的点则无解，找到入度为0的点作为根进行dfs，因为是一条链，如果出现分叉点（新图的点的邻接边数目>=2）则无解。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;

const int maxn=1005;

vector<int> G[maxn];
vector<int> rG[maxn];
vector<int> nG[maxn];
vector<int> vs;

int n,m;

int cmp[maxn],d[maxn];
bool used[maxn],tag;

inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
    while(ch<='9'&&ch>='0'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

void init(){
    for(int i=0;i<maxn;i++){
        G[i].clear();rG[i].clear();nG[i].clear();
        d[i]=0;cmp[i]=0;
    }
    tag=false;
}

void add_edge(int s,int t){
    G[s].push_back(t);
    rG[t].push_back(s);
}

void dfs(int s){
    used[s]=true;
    for(int i=0;i<G[s].size();i++){
        if(!used[G[s][i]]) dfs(G[s][i]);
    }
    vs.push_back(s);
}

void rdfs(int s,int k){
    used[s]=true;
    cmp[s]=k;
    for(int i=0;i<rG[s].size();i++){
        if(!used[rG[s][i]]) rdfs(rG[s][i],k);
    }
}

void ndfs(int root){
    if(nG[root].size()>1){
        tag=true;
        return;
    }
    if(nG[root].size()==0) return;
    ndfs(nG[root][0]);
}

int scc(){
    memset(used,0,sizeof(used));
    vs.clear();
    for(int i=1;i<=n;i++){
        if(!used[i]) dfs(i);
    }
    memset(used,0,sizeof(used));
    int k=0;
    for(int i=vs.size()-1;i>=0;i--){
        if(!used[vs[i]]) rdfs(vs[i],++k);
    }
    return k;
}

int main(){
    int T;
    T=read();
    while(T--){
        init();
        n=read();m=read();
        for(int i=0;i<m;i++){
            int s,t;
            s=read();t=read();
            add_edge(s,t);
        }
        int k=scc();
        for(int i=1;i<=n;i++){
            for(int j=0;j<G[i].size();j++){
                int v=G[i][j];
                if(cmp[i]==cmp[v]) continue;
                nG[cmp[i]].push_back(cmp[v]);
                d[cmp[v]]++;
            }
        }
        int root=0;bool flag=false;
        for(int i=1;i<=k;i++){
            if(d[i]==0){
                if(root){
                    flag=true;
                    break;
                }
                root=i;
            }
        }
        if(flag){
            printf("No\n");
        }
        else{
            ndfs(root);
            if(!tag) printf("Yes\n");
            else printf("No\n");
        }
    }
    return 0;
}