POJ-2762-Going from u to v or from v to u（强连通， 拓扑排序）

链接：

https://vjudge.net/problem/POJ-2762

题意：

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?
给一个有向图，求判断能否选任意的两个点，有一条从u到v或v到u的路径。

思路：

先缩点形成一个DAG。只有当这个DAG是一条链的时候，才有u到v的路径。

代码：

#include <iostream>
#include <cstdio>
#include <vector>
#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
using namespace std;
const int MAXN = 1e3+10;

stack<int> St;
vector<int> G_new[MAXN], G[MAXN];
int Vis[MAXN];
int Dfn[MAXN], Low[MAXN];
int Fa[MAXN];
int Dis[MAXN];
int n, m;
int times, cnt;

void Tarjan(int x)
{
    Dfn[x] = Low[x] = ++times;
    Vis[x] = 1;
    St.push(x);
    for (int i = 0;i < G[x].size();i++)
    {
        int node = G[x][i];
        if (Dfn[node] == 0)
        {
            Tarjan(node);
            Low[x] = min(Low[x], Low[node]);
        }
        else if (Vis[node])
            Low[x] = min(Low[x], Dfn[node]);
    }
    if (Dfn[x] == Low[x])
    {
        cnt++;
        while (St.top() != x)
        {
            Fa[St.top()] = cnt;
            Vis[St.top()] = 0;
            St.pop();
        }
        Fa[St.top()] = cnt;
        Vis[St.top()] = 0;
        St.pop();
    }
}

void Init()
{
    for (int i = 1;i <= n;i++)
        G[i].clear(), G_new[i].clear();
    memset(Vis, 0, sizeof(Vis));
    memset(Dis, 0, sizeof(Dis));
    memset(Dfn, 0, sizeof(Dfn));
    times = cnt = 0;
    while (St.size())
        St.pop();
}

bool Tupo()
{
    int sum = 0;
    stack<int> tu;
    for (int i = 1;i <= cnt;i++)
        if (Dis[i] == 0)
            tu.push(i);
    while (!tu.empty())
    {
        if (tu.size() > 1)
            return false;
        int x = tu.top();
        sum++;
        tu.pop();
        for (int i = 0;i < G_new[x].size();i++)
        {
            int node = G_new[x][i];
            if (--Dis[node] == 0)
                tu.push(node);
        }
    }
    return sum == cnt;
}

int main()
{
    int t;
    int l, r;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d%d", &n, &m);
        Init();
        for (int i = 1;i <= m;i++)
        {
            scanf("%d%d", &l, &r);
            G[l].push_back(r);
        }
        for (int i = 1;i <= n;i++)
            if (Dfn[i] == 0)
                Tarjan(i);
        for (int i = 1;i <= n;i++)
        {
            for (int j = 0;j < G[i].size();j++)
            {
                int node = G[i][j];
                if (Fa[i] != Fa[node])
                {
                    Dis[Fa[node]]++;
                    G_new[Fa[i]].push_back(Fa[node]);
                }
            }
        }
        if (Tupo())
            printf("Yes
");
        else
            printf("No
");
    }

    return 0;
}