POJ 2762 Going from u to v or from v to u? （图论-tarjan）

Going from u to v or from v to u?

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14469		Accepted: 3801

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

题目大意：

T组测试数据，n个点，m有向条边，然后问你这个图是否满足任意两个点u到v或者v到u

解题思路：

参照我的博客这篇文章，就知道解法了：/article/1615777.html

解题代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;

const int maxn=1100;
const int maxm=51000;

struct edge{
    int u,v,next;
    edge(int u0=0,int v0=0){
        u=u0;v=v0;
    }
}e[maxm];

int n,m,head[maxn],dfn[maxn],low[maxn],mark[maxn],w[maxn],color[maxn],dp[maxn],cnt,nc,index;
vector <int> vec;
vector <vector<int> > dfsmap;

void addedge(int u,int v){
    e[cnt]=edge(u,v);e[cnt].next=head[u];head[u]=cnt++;
}

void input(){
    cnt=nc=index=0;
    scanf("%d%d",&n,&m);
    vec.clear();
    for(int i=0;i<=n;i++){
        w[i]=dfn[i]=0;
        mark[i]=false;
        color[i]=dp[i]=head[i]=-1;
    }
    int u,v;
    while(m-- >0){
        scanf("%d%d",&u,&v);
        addedge(u,v);
    }
}

void tarjan(int s){
    dfn[s]=low[s]=++index;
    mark[s]=true;
    vec.push_back(s);
    for(int i=head[s];i!=-1;i=e[i].next){
        int d=e[i].v;
        if(!dfn[d]){
            tarjan(d);
            low[s]=min(low[d],low[s]);
        }else if(mark[d]){
            low[s]=min(low[s],dfn[d]);
        }
    }
    if(dfn[s]==low[s]){
        nc++;
        int d;
        do{
            d=vec.back();
            vec.pop_back();
            color[d]=nc;
            mark[d]=false;
            w[nc]++;
        }while(d!=s);
    }
}

int DP(int s){
    if(dp[s]!=-1) return dp[s];
    int ans=w[s];
    for(int i=0;i<dfsmap[s].size();i++){
        int d=dfsmap[s][i];
        if(DP(d)+w[s]>ans) ans=DP(d)+w[s];
    }
    return dp[s]=ans;
}

void solve(){
    for(int i=1;i<=n;i++){
        if(!dfn[i]) tarjan(i);
    }
    dfsmap.clear();
    dfsmap.resize(nc+1);
    for(int i=0;i<cnt;i++){
        int x=color[e[i].u],y=color[e[i].v];
        if(x!=y){
            dfsmap[x].push_back(y);
            //cout<<x<<"->"<<y<<endl;
        }
    }
    int ans=0;
    for(int i=1;i<=nc;i++){
        if(DP(i)>ans) ans=DP(i);
        //cout<<i<<" "<<ans<<endl;
    }
    if(ans>=n) printf("Yes\n");
    else printf("No\n");
}

int main(){
    int t;
    scanf("%d",&t);
    while(t-- >0){
        input();
        solve();
    }
    return 0;
}