牛客-15705 可达性

可达性

\(tarjan\) 缩点

如果是 \(DAG\) 图，则答案就是入度为 \(0\) 的强连通分量中字典序最小的点组成的点集

因此直接缩点，然后维护每个强连通分量的最小字典序的点

#include <iostream>
#include <cstdio>
#include <stack>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn = 1e5 + 10;
vector<int>gra[maxn];
int low[maxn], dfn[maxn], tp = 0;
int vis[maxn], du[maxn];
int scc[maxn], cnt_scc = 0;
int minn[maxn];
stack<int>st;

void tarjan(int now)
{
    st.push(now);
    vis[now] = 1;
    dfn[now] = low[now] = ++tp;
    for(int nex : gra[now])
    {
        if(dfn[nex] == 0)
        {
            tarjan(nex);
            low[now] = min(low[nex], low[now]);
        }
        else if(vis[nex])
            low[now] = min(low[nex], low[now]);
    }
    if(low[now] == dfn[now])
    {
        cnt_scc++;
        minn[cnt_scc] = maxn + 10;
        while(st.top() != now)
        {
            int x = st.top();
            st.pop();
            vis[x] = 0;
            scc[x] = cnt_scc;
            minn[cnt_scc] = min(minn[cnt_scc], x);
        }
        st.pop();
        vis[now] = 0;
        scc[now] = cnt_scc;
        minn[cnt_scc] = min(minn[cnt_scc], now);
    }
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n, m;
    cin >> n >> m;
    for(int i=0; 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 nex : gra[i])
            if(scc[i] != scc[nex])
                du[scc[nex]]++;
    int cur = 0, ans = 0;
    for(int i=1; i<=cnt_scc; i++) if(du[i] == 0) ans++;
    cout << ans << endl;
    for(int i=1; i<=cnt_scc; i++)
    {
        if(du[i] == 0)
        {
            if(cur++) cout << " ";
            cout << minn[i];
        }
    }
    cout << endl;
}