hdu 4635 Strongly connected（连通分量）

#include 
#include 
#include 
using namespace std;
/**
	在原有的有向连通图中插入尽可能多的边使其不是强连通 
	由于原图非强连通，因此至少存在一个强连通分支入度或出度为0。而且最后答案中的图中
	也应该存在这样的分支。通过添加边无法减少度，因此需要从度为0的分支中找出点数最少的并
	使其与外部的点相连且保留原属性 
**/
typedef __int64 LL;
const int MAXN = 100005;
int n, m;
int head[MAXN], en;
int dfn[MAXN], low[MAXN], cnt;
int isin[MAXN], mstk[MAXN], top;
int cfz[MAXN], nfz, fz[MAXN];
int id[MAXN], od[MAXN];
struct edge
{
	int v, next;
}e[MAXN];
void add(int u, int v)
{
	e[en].v = v; e[en].next = head[u]; head[u] = en++;
}

void solve(int u)
{
	dfn[u] = low[u] = ++cnt;
	isin[u] = 1;
	mstk[++top] = u;
	for (int i = head[u]; ~i; i = e[i].next)
	{
		int v = e[i].v;
		if (!dfn[v])
		{
			solve(v);
			low[u] = min(low[v], low[u]);
		}
		else if (isin[v])
		{
			low[u] = min(low[u], dfn[v]);
		}
	}
	if (dfn[u] == low[u])
	{
		while (1)
		{
			int v = mstk[top];
			isin[v] = 0;
			cfz[nfz]++, --top;
			fz[v] = nfz;
			if (u == v) break;
		}
		nfz++;
	}
}
void minit()
{
	memset(head, -1, sizeof head);
	en = 0;
	memset(dfn, 0, sizeof dfn); memset(low, 0, sizeof low);
	cnt = 0;
	memset(isin, 0, sizeof isin); top = 0;
	nfz = 0;
	memset(cfz, 0, sizeof cfz);	
	memset(id, 0, sizeof id); memset(od, 0, sizeof od);
}
int main()
{
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
#endif
	int t, cs = 0;
	scanf("%d", &t);
	while (t--)
	{
		printf("Case %d: ", ++cs);
		scanf("%d%d", &n, &m);
		minit();
		for (int i = 0; i< m; ++i)
		{
			int u, v;
			scanf("%d%d", &u, &v);
			add(u, v);
		}
		for (int i= 1; i<= n; ++i)
		{
			if (!dfn[i]) solve(i);
		}
		if (nfz == 1)
		{
			printf("-1\n"); continue;
		}
		for (int u = 1; u<= n; ++u)
		{
			for (int i = head[u]; ~i; i = e[i].next)
			{
				int v = e[i].v;
				if (fz[u] != fz[v])
				{
					od[fz[u]]++; id[fz[v]]++;
				}
			}
		}
		LL sum = -1;
		for (int i = 0; i< nfz; ++i)
		{
			LL x = cfz[i], y = n-cfz[i];
			if (id[i] == 0 || od[i] == 0)
			{
				sum = max(sum, x*(x-1) + y*(y-1) 
				+ x*(n-x) - m);
			}
		}
		printf("%I64d\n", sum);
	}
	return 0;
}