POJ 2762 Going from u to v or from v to u?

Going from u to v or from v to u?

Time Limit: 2000ms

Memory Limit: 65536KB

This problem will be judged on PKU. Original ID: 2762
64-bit integer IO format: %lld      Java class name: Main

 

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

 

解题：强连通缩点+拓扑排序。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 1005;
struct arc{
    int to,next;
    arc(int x = 0,int y = -1){
        to = x;
        next = y;
    }
};
int dfn[maxn],low[maxn],belong[maxn],tot;
int head[maxn],head2[maxn],in[maxn],scc,cnt;
stack<int>stk;
bool instack[maxn];
arc e[500000];
void add(int *hd,int u,int v){
    e[tot] = arc(v,hd[u]);
    hd[u] = tot++;
}
void tarjan(int u){
    dfn[u] = low[u] = ++cnt;
    stk.push(u);
    instack[u] = true;
    for(int i = head[u]; ~i; i = e[i].next){
        if(!dfn[e[i].to]){
            tarjan(e[i].to);
            if(low[e[i].to] < low[u]) low[u] = low[e[i].to];
        }else if(instack[e[i].to] && dfn[e[i].to] < low[u]) low[u] = dfn[e[i].to];
    }
    if(dfn[u] == low[u]){
        scc++;
        int v;
        do{
           v = stk.top();
           stk.pop();
           instack[v] = false;
           belong[v] = scc;
        }while(v != u);
    }
}
bool topsort(){
    queue<int>q;
    for(int i = 1; i <= scc; i++){
        if(in[i]) continue;
        q.push(i);
    }
    if(q.size() > 1) return false;
    while(!q.empty()){
        int u = q.front();
        q.pop();
        for(int i = head2[u]; ~i; i = e[i].next)
            if(--in[e[i].to] == 0) q.push(e[i].to);
        if(q.size() > 1) return false;
    }
    return true;
}
int main() {
    int t,u,v,n,m;
    scanf("%d",&t);
    while(t--){
        scanf("%d %d",&n,&m);
        for(int i = 0; i < maxn; i++){
            dfn[i] = low[i] = 0;
            instack[i] = false;
            in[i] = belong[i] = 0;
            head[i] = head2[i] = -1;
        }
        scc = cnt = tot = 0;
        while(!stk.empty()) stk.pop();
        for(int i = 0; i < m; i++){
            scanf("%d %d",&u,&v);
            add(head,u,v);
        }
        for(int i = 1; i <= n; i++)
            if(!dfn[i]) tarjan(i);
        if(scc == 1) {puts("Yes");continue;}
        for(int i = 1; i <= n; i++){
            for(int j = head[i]; ~j; j = e[j].next){
                if(i == e[j].to || belong[i] == belong[e[j].to]) continue;
                in[belong[e[j].to]]++;
                add(head2,belong[i],belong[e[j].to]);
            }
        }
        if(topsort()) puts("Yes");
        else puts("No");
    }
    return 0;
}