POJ 1236.Network of Schools 解题报告

/*
       无向图强连通的Garbow算法，思路与Tarjan算法相同，实现更直接，效率更好
       时间复杂度同样为O（n+m）
       思路：
       dfn记录访问顺序，st为访问栈，tem为辅助栈
       每次找到环时，将环中除顺序最靠前的点其他的点全出栈st
       tem中当前点之上（包括当前点）的所有点为一个强连通分量
*/
#include <iostream>
#include <cstring>
using namespace std;
const int INF = 109;
struct node {
	int u, v, ne;
} E[INF*INF];
int head[INF], cnt;
int dfn[INF], num[INF], sta[INF], Tops, tem[INF], Topt, scc;
int n;
void addedge (int u, int v) {
	E[++cnt].u = u, E[cnt].v = v;
	E[cnt].ne = head[u];
	head[u] = cnt;
}
void dfs (int k, int t) {
	sta[++Tops] = tem[++Topt] = k;
	dfn[k] = ++t;
	for (int i = head[k]; i != 0; i = E[i].ne) {
		int v = E[i].v;
		if (!dfn[v]) dfs (v, t);
		else if (num[v] == 0)
			while (dfn[sta[Tops]] > dfn[v]) Tops--;
	}
	if (sta[Tops] == k) {
		Tops--, scc++;
		do
			num[tem[Topt]] = scc;
		while (tem[Topt--] != k);
	}
}
void Garbow (int n) {
	memset (dfn, 0, sizeof dfn);
	memset (num, 0, sizeof num);
	Tops = Topt = scc = 0;
	for (int i = 1; i <= n; i++)
		if (!num[i]) dfs (i, 0);
}
void make() {
	Garbow (n);
	int degi[INF], dego[INF];
	memset (degi, 0, sizeof degi);
	memset (dego, 0, sizeof dego);
	if (scc == 1) {
		cout << 1 << endl << 0 << endl;
		return ;
	}
	for (int i = 1; i <= cnt; i++) {
		int u = E[i].u, v = E[i].v;
		if (num[u] != num[v]) {
			degi[num[v]] = 1;
			dego[num[u]] = 1;
		}
	}
	int ans1=0, ans2=0;
	for (int i = 1; i <= scc; i++) {
		if (degi[i] == 0) ans1++;
		if (dego[i] == 0) ans2++;
	}
	cout << ans1 <<endl<< max (ans1, ans2);
}
int main() {
	cin >> n;
	for (int i = 1; i <= n; i++) {
		int x;
		while (cin >> x && x)
			addedge (i, x);
	}
	make();
	return 0;
}