Poj 2762 going from u to V or from V to u? (Graph theory-Tarjan)

Going from u to V or from V to u?

Time limit:2000 ms		Memory limit:65536 K
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 Po Ssible. 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 shoshould contain t lines. Write 'yes' if the cave has the property stated above, or 'no' otherwise.

Sample Input

13 31 22 33 1

Sample output

Yes

Source

Poj monthly -- 2006.02.26, zgl & twb

Question:

T group test data, n vertices, M has a directed edge, and then asks if this graph meets any two vertices u to V or V to u

Solution:

Refer to my blog this article, you will know the solution: http://blog.csdn.net/a1061747415/article/details/38380665

Solution code:

#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;}