POJ 2762 Going from u to v or from v to u? 强连通+判断链

Going from u to v or from v to u?

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 13963		Accepted: 3640

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

#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <list>
#include <stdexcept>
#include <functional>
#include <utility>
#include <ctime>

using namespace std;

#define PB push_back
#define MP make_pair
#define REP(i,n) for(int i=0;i<(n);++i)
#define FOR(i,l,h) for(int i=(l);i<=(h);++i)
#define DWN(i,h,l) for(int i=(h);i>=(l);--i)
#define CLR(vis) memset(vis,0,sizeof(vis))
#define MST(vis,pos) memset(vis,pos,sizeof(vis))
#define MAX3(a,b,c) max(a,max(b,c))
#define MAX4(a,b,c,d) max(max(a,b),max(c,d))
#define MIN3(a,b,c) min(a,min(b,c))
#define MIN4(a,b,c,d) min(min(a,b),min(c,d))
#define PI acos(-1.0)
#define INF 0x7FFFFFFF
#define LINF 1000000000000000000LL
#define eps 1e-8

typedef long long ll;

const int maxn=1111;

int g[maxn][maxn];
int in[maxn];

struct node{
    int u,to,next,w;
}e[maxn*maxn];

int id[maxn],q[maxn],dfn[maxn],low[maxn],num[maxn];

int n,m,tsp,qe,cnt,edge,head[maxn];

void init()
{
    memset(head,-1,sizeof(head));
    edge=0;
}

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

void tarjan(int u)
{
    int i,v;
    dfn[u]=low[q[qe++]=u]=++tsp;
    for(i=head[u];i>=0;i=e[i].next)
        if(!dfn[v=e[i].to])
        {
            tarjan(v);
            low[u]=min(low[u],low[v]);
        }
        else
        {
            if(id[v]<0)
            {
              low[u]=min(low[u],dfn[v]);
            }
        }
    if(low[u]==dfn[u])
    {
        ++cnt;
        do{
             v=q[--qe];
             ++num[id[v]=cnt];

        }while(v!=u);
    }
}

void solve()
{
    int i;
    tsp=qe=cnt=0;
    for(i=0;i<=n;++i)
        id[i]=-1,dfn[i]=0;
    for(i=1;i<=n;++i)
        if(!dfn[i])
            tarjan(i);
}

bool check()
{
    queue<int> q;
    FOR(i,1,cnt)
    {
        if(!in[i])
            q.push(i);
    }
    if(q.size()>1)
        return false;
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        FOR(i,1,cnt)
        {
            if(g[u][i])
            {
                in[i]--;
               if(!in[i])
                q.push(i);
            }
        }
        if(q.size()>1)
            return false;
    }
    return true;
}

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        init();
        int u,v;
        scanf("%d%d",&n,&m);
        REP(i,m)
        {
            scanf("%d%d",&u,&v);
            addedge(u,v);
        }

        solve();

        if(cnt==1)
        {
            printf("Yes\n");
            continue;
        }

        CLR(g);

        CLR(in);

        FOR(u,1,n)
        {
            for(int j=head[u];~j;j=e[j].next)
            {
                int v=e[j].to;
                if(u!=v && id[u]!=id[v])
                {
                   g[id[u]][id[v]]=1;
                   in[id[v]]++;
                }
            }
        }
        if(check())
            printf("Yes\n");
        else
            printf("No\n");

    }
    return 0;
}