【luogu P2002】消息扩散

https://www.luogu.org/problem/show?pid=2002

SCC缩点的模板题，缩点后统计入度为0的点的数量就完了。

#include <iostream>
#include <vector>
#include <stack>
#include <cstring>
#define maxn 100005
using namespace std;
int n, m;
vector<int> g[maxn], g2[maxn];
int indegree[maxn];

int timer = 0, cnter = 0;
int dfn[maxn], low[maxn], scc[maxn];
stack<int> sta;
bool insta[maxn];
void dfs(int v)
{
    /*
    low[v]=min{
        dfn[v],
        low[w], 存在有向边(v,w)且搜索到v时w尚未被搜索到
        dfn[w]  存在有向边(v,w)且搜索到v时w在栈内
    }
    */
    low[v] = dfn[v] = ++timer;
    sta.push(v);
    insta[v] = true;
    for (int i = 0; i < g[v].size(); i++)
    {
        int w = g[v][i];
        if (!dfn[w])
        {
            dfs(w);
            low[v] = min(low[v], low[w]);
        }
        else if (insta[w])
        {
            low[v] = min(low[v], dfn[w]);
        }
    }
    if (dfn[v] == low[v])
    {
        ++cnter;
        int t;
        do
        {
            t = sta.top();
            sta.pop();
            insta[t] = false;
            scc[t] = cnter;
        } while (t != v);
    }
}
void tarjan_scc()
{
    for (int i = 1; i <= n; i++)
    {
        if (!dfn[i])
            dfs(i);
    }
}
void contract()
{
    for (int v = 1; v <= n; v++)
    {
        for (int i = 0; i < g[v].size(); i++)
        {
            // 对于边(v,w)，若v与w不在同一SCC，则该边在缩点后的新图为(scc[v],scc[w])
            int w = g[v][i];
            if (scc[v] != scc[w])
            {
                g2[scc[v]].push_back(scc[w]);
                indegree[scc[w]]++;
            }
        }
    }
}
int main()
{
    ios::sync_with_stdio(false);
    cin >> n >> m;
    int u, v;
    for (int i = 1; i <= m; i++)
    {
        cin >> u >> v;
        g[u].push_back(v);
    }
    tarjan_scc();
    contract();

    // 统计得到的DAG中入度为0的点
    int ans = 0;
    for (int i = 1; i <= cnter; i++)
    {
        if (!indegree[i])
            ans++;
    }
    cout << ans << endl;
    return 0;
}