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https://www.luogu.org/problemnew/show/2746

Tarjan 缩点

判断入度为0的点的个数与出度为0的点的个数的关系

注意全缩为一个点的情况

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
 
using namespace std;
const int N = ;
 
#define yxy getchar()
 
int n, now = , Tarjan_tim, Bel_tim, Top;
int head[N], Low[N], Dfn[N], Out[N], In[N], Belong[N], Stack[N];
bool vis[N];
struct Node {int u, v, nxt;} G[N * N * ];
 
inline int read(){
    int x = ; char c = yxy;
    while(c < '' || c > '') c = yxy;
    while(c >= '' && c <= '') x = x *  + c - '', c = yxy;
    return x;
}
 
inline void add(int u, int v){
    G[now].v = v; G[now].nxt = head[u]; head[u] = now ++;
}
 
void Tarjan(int u){
    Dfn[u] = Low[u] = ++ Tarjan_tim;
    vis[u] = ; Stack[++ Top] = u;
    for(int i = head[u]; ~ i; i = G[i].nxt){
        int v = G[i].v;
        if(!Dfn[v]){
            Tarjan(v);
            Low[u] = min(Low[u], Low[v]);
        }
        else if(vis[v]) Low[u] = min(Low[u], Dfn[v]);
    }
    if(Dfn[u] == Low[u]){
        Bel_tim ++;
        Belong[u] = Bel_tim;
        vis[u] = ;
        while(Stack[Top] != u){
            Belong[Stack[Top]] = Bel_tim;
            vis[Stack[Top]] = ;
            Top --;
        }   Top --;
    }
}
 
void Build_G(){
    for(int u = ; u <= n; u ++){
        for(int i = head[u]; ~ i; i = G[i].nxt){
            int v = G[i].v;
            if(Belong[u] != Belong[v]){
                Out[Belong[u]] ++; In[Belong[v]] ++;
            }
        }
    }
}
 
int main()
{
    n = read();
    for(int i = ; i <= n; i ++) head[i] = -;
    for(int i = ; i <= n; i ++){
        int v = read();
        while(v) {
            add(i, v);
            v = read();
        }
    }
    for(int i = ; i <= n; i ++) if(!Dfn[i]) Tarjan(i);
    Build_G();
    int Out_js(), In_js();
    for(int i = ; i <= Bel_tim; i ++){
        if(!Out[i]) Out_js ++;
        if(!In[i])  In_js  ++;
    }
    if(Bel_tim == ) cout << "" << "\n" << "";
    else cout << In_js << "\n" << max(Out_js, In_js);
    return ;
}