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

Going from u to v or from v to u?

判断是不是弱连通图

\(tarjan\) 缩点 + 拓扑

弱联通说明必须要有唯一的拓扑序，也就是拓扑的队列里始终只有一个

拓扑需要 \(DAG\) 图，因此考虑 \(tarjan\) 缩点后跑拓扑

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <stack>
#include <vector>
#include <queue>
using namespace std;
const int maxn = 1010;
vector<int>gra[maxn], gra_c[maxn];
int low[maxn], dfn[maxn], tp = 0;
int scc[maxn], cnt_scc = 0;
int vis[maxn], du[maxn];
stack<int>st;

void tarjan(int now)
{
    vis[now] = 1;
    st.push(now);
    dfn[now] = low[now] = ++tp;
    for(int i=0; i<gra[now].size(); i++)
    {
        int nex = gra[now][i];
        if(dfn[nex] == 0)
        {
            tarjan(nex);
            low[now] = min(low[now], low[nex]);
        }    
        else if(vis[nex])
            low[now] = min(low[now], low[nex]);
    }
    if(low[now] == dfn[now])
    {
        cnt_scc++;
        while(st.top() != now)
        {
            int x = st.top();
            vis[x] = 0;
            scc[x] = cnt_scc;
            st.pop();
        }
        vis[now] = 0;
        scc[now] = cnt_scc;
        st.pop();
    }
}

void init(int n)
{
    for(int i=0; i<=n; i++) low[i] = dfn[i] = scc[i] = du[i] = 0;
    tp = cnt_scc = 0;
    for(int i=0; i<=n; i++) {gra[i].clear(); gra_c[i].clear();}
}

bool topu()
{
    queue<int>q;
    int cnt = 0;
    for(int i=1; i<=cnt_scc; i++)
        if(du[i] == 0) q.push(i);
    while(q.size())
    {
        if(q.size() != 1) return false;
        int now = q.front();
        q.pop();
        cnt++;
        for(int i=0; i<gra_c[now].size(); i++)
        {
            int nex = gra_c[now][i];
            if(--du[nex] == 0)
                q.push(nex);
        }
    }
    return cnt == cnt_scc;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while(t--)
    {
        int n, m;
        cin >> n >> m;
        init(n);
        for(int i=1; i<=m; i++)
        {
            int a, b;
            cin >> a >> b;
            gra[a].push_back(b);
        }
        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<gra[i].size(); j++)
            {
                int nex = gra[i][j];
                if(scc[i] != scc[nex])
                {
                    gra_c[scc[i]].push_back(scc[nex]);
                    du[scc[nex]]++;
                }
            }
        }
        if(topu()) cout << "Yes\n";
        else cout << "No\n";
    }
    
}