HDU 1236 强连通分量

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn = 110;
const int maxm = maxn * maxn;
struct Edge
{
	int to, next;
};
Edge edge[maxm];
int head[maxn], tol, v, n, low[maxn], DFN[maxn], Stack[maxn], Belong[maxn], index, top, scc;
bool Instack[maxn];
void addedge(int u, int v)
{
	edge[tol].to = v;
	edge[tol].next = head[u];
	head[u] = tol++;
}
void Tarjan(int u)
{
	int v;
	low[u] = DFN[u] = ++index;
	Stack[top++] = u;
	Instack[u] = true;
	for (int i = head[u]; i != -1; i = edge[i].next)
	{
		v = edge[i].to;
		if (!DFN[v])
		{
			Tarjan(v);
			if (low[u] > low[v])
				low[u] = low[v];
		}
		else if (Instack[v] && low[u] > DFN[v])
			low[u] = DFN[v];
	}
	if (low[u] == DFN[u])
	{
		scc++;
		do
		{
			v = Stack[--top];
			Belong[v] = scc;
			Instack[v] = false;
		}
		while (v != u);
	}
}
int in[maxn], out[maxn];
void solve(int N)
{
	memset(DFN, 0, sizeof(DFN));
	memset(Instack, 0, sizeof(Instack));
	index = scc = top = 0;
	for (int i = 1; i <= N; i++)
		if (!DFN[i]) Tarjan(i);
	if (scc == 1) {printf("1\n0\n"); return;}
	for (int i = 1; i <= scc; i++)
		in[i] = out[i] = 0;
	for (int u = 1; u <= N; u++)
		for (int i = head[u]; i != -1; i = edge[i].next)
		{
			int v = edge[i].to;
			if (Belong[u] != Belong[v])
				in[Belong[v]]++, out[Belong[u]]++;
		}
	int ans1 = 0, ans2 = 0;
	for (int i = 1; i <= scc; i++)
	{
		if (in[i] == 0) ans1++;
		if (out[i] == 0) ans2++;
	}
	printf("%d\n%d\n", ans1, max(ans1, ans2));
}
int main(int argc, char const *argv[])
{
	while (~scanf("%d", &n))
	{
		tol = 0;
		memset(head, -1, sizeof(head));
		for (int i = 1; i <= n; i++)
			while (~scanf("%d", &v) && v)
				addedge(i, v);
		solve(n);
	}
	return 0;
}

这题给了一个有向图。需要解决两个问题：
第一是需要给多少个点，才能传遍所有点。
第二问是加多少条边，使得整个图变得强连通。
使用Tarjan进行缩点，得到一个SCC图，统计有n个入度为零，m个出度为零的；

问题一答案为n， 问题二答案为max(n,m)；