poj2762 Going from u to v or from v to u? 有向图 强连通分量 拓扑排序

Going from u to v or from v to u?

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 13040		Accepted: 3383

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

Source

POJ Monthly--2006.02.26,zgl & twb

这题要注意,边是单向边,先用强连通分量缩点,建图,再拓扑排序若是单向链刚对,否刚为错就可以a了!

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define N 1051
#define E 30000
#define M 1000000
using namespace std;
int head[N],head2[N],next2[E],vec2[E],sta[M],re,ans,next[E],id[N],in[N],vec[E],vis[N],dfn[N],low[N],clock_m,edge_m,edge_m2;
int addedge(int s,int e){
    vec[edge_m]=e;next[edge_m]=head[s];head[s]=edge_m++;
}
int addedge2(int s,int e){
    vec2[edge_m2]=e;next2[edge_m2]=head2[s];head2[s]=edge_m2++;
}
int init(){
    memset(vis,0,sizeof(vis));
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(head,-1,sizeof(head));
    memset(head2,-1,sizeof(head2));
    memset(id,0,sizeof(id));
    memset(in,0,sizeof(in));
    edge_m=0;clock_m=0;ans=0;re=0;edge_m2=0;
}
int tarjan(int x){
    dfn[x]=low[x]=++clock_m;
    sta[++ans]=x;
    vis[x]=1;
    for(int i=head[x];i!=-1;i=next[i]){
        int goal=vec[i];
        if(!dfn[goal]){
            tarjan(goal);
            low[x]=min(low[x],low[goal]);
        }
        else if(/*vis[goal]*/!id[goal])
            low[x]=min(low[x],dfn[goal]);
    }
    if(low[x]==dfn[x]){
        re++;int v;
       do{
            v=sta[ans--];
            vis[v]=0;
            id[v]=re;
        }while(v!=x);
    }
    return 1;
}
int topsort(int n){
    ans=0;
    for(int i=1;i<=re;i++){
        if(in[i]==0){
            sta[ans++]=i;
        }
    }
    if(ans>1)return 0;
    while(ans>0){
        ans--;
        int qtop=sta[ans];
        for(int j=head2[qtop];j!=-1;j=next2[j]){
            in[vec2[j]]--;
            if(in[vec2[j]]==0)
            sta[ans++]=vec2[j];
        }
        if(ans>1)
        return 0;
    }
    return 1;
}
int main()
{
    int tcase,n,m,s,e;
    scanf("%d",&tcase);
    while(tcase--){
        //system("PAUSE");
        init();
        scanf("%d%d",&n,&m);
        for(int i=0;i<m;i++){
            scanf("%d%d",&s,&e);
            addedge(s,e);
        }
        for(int i=1;i<=n;i++)
        if(!dfn[i])
        {
           tarjan(i);
        }
       if(re==1){
            printf("Yes\n");
            continue;
        }
        for(int i=1;i<=n;i++){
            for(int j=head[i];j!=-1;j=next[j]){
                if(id[vec[j]]!=id[i]){
                    in[id[vec[j]]]++;
                    addedge2(id[i],id[vec[j]]);
                }
            }
        }
        if(topsort(re))printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}