【题解】LuoGu2746：校园网Network of Schools

原题传送门
又是可以用上强连通分量
使用tarjan缩点，缩好的点建新图

问1：就是求新图入度为0的点个数，没毛病吧
问2：任一点都能到达其他点，说明是形成了一个环。环中任意一点入度出度不为0，那么我们只要求出入度为0的点个数与出度为0的点个数的更大的那个就行了（yy一下）

有一个注意点就是，新图中只有一个点的情况要特判，答案是$1$ $0$

Code：

#include <bits/stdc++.h>
#define maxn 10010
using namespace std;
struct Edge{
    int to, next;
}edge[maxn << 1];
struct Line{
    int x, y;
}line[maxn];
int num, head[maxn], dfn[maxn], low[maxn], Index, tot;
int top, cnt, sta[maxn], vis[maxn], n, color[maxn], in[maxn], out[maxn];

inline int read(){
    int s = 0, w = 1;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
    for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
    return s * w;
}

void addedge(int x, int y){ edge[++num] = (Edge) { y, head[x] }; head[x] = num; }

void tarjan(int u){
    dfn[u] = low[u] = ++Index;
    vis[u] = 1, sta[++top] = u;
    for (int i = head[u]; i; i = edge[i].next){
        int v = edge[i].to;
        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]){
        ++tot;
        while (sta[top + 1] != u)
            color[sta[top]] = tot, vis[sta[top--]] = 0;
    }
}

int main(){
    n = read();
    for (int i = 1; i <= n; ++i){
        int x = read();
        while (x) line[++cnt] = (Line) { i, x }, addedge(i, x), x = read();
    }
    for (int i = 1; i <= n; ++i)
        if (!dfn[i]) tarjan(i);
    num = 0;
    memset(head, 0, sizeof(head));
    for (int i = 1; i <= cnt; ++i)
        if (color[line[i].x] != color[line[i].y])
            addedge(color[line[i].x], color[line[i].y]),
            ++in[color[line[i].y]], ++out[color[line[i].x]];
    int ans1 = 0, ans2 = 0;
    for (int i = 1; i <= tot; ++i){
        if (!in[i]) ++ans1;
        if (!out[i]) ++ans2;
    }
    if (tot == 1) printf("1\n0\n"); else
    printf("%d\n%d\n", ans1, max(ans1, ans2));
    return 0;
}