SPOJ-CAPCITY Capital City

Capital City

tarjan 缩点

缩点之后，找到 DAG 图中唯一一个出度为 \(0\) 的点，如果有多个，说明不成立

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
using namespace std;
const int maxn = 2e5 + 10;
vector<int>gra[maxn], gra_c[maxn];
int low[maxn], dfn[maxn], tp = 0;
int scc[maxn], cnt_scc = 0, vis[maxn];
int cnt[maxn], du[maxn];
stack<int>st;

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

void build(int n)
{
    for(int i=1; i<=n; i++)
    {
        for(int nex : gra[i])
        {
            if(scc[i] != scc[nex])
            {
                du[scc[i]]++;
            }
        }
    }
}

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);
    build(n);
    int x = 0, way = -1;
    for(int i=1; i<=cnt_scc; i++)
        if(du[i] == 0) {x++; way = i;}
    if(x != 1) way = 0;
    cout << cnt[way] << endl;
    int cur = 0;
    for(int i=1; i<=n; i++)
    {
        if(scc[i] == way)
        {
            if(cur++) cout << " ";
            cout << i;
        }
    }
    cout << endl;
    return 0;
}