ZOJ 3630 —— Information

原題：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3630

題意：有n個點，m條邊，刪除一個點後，得到強連通分量點數的最大值，求最大點數的最小值；

注意：點數 = 1 時爲 0 ；

思路：枚舉每個點爲刪除點，求出每次強連通分量的最大值，然後取最小的；

#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<algorithm>
#define inf 0x3f3f3f3f
using namespace std;
const int maxn = 110;
const int maxm = 10000;

int n, m;
int DFN[maxn], Low[maxn], Stack[maxn];
bool Instack[maxn];
int Time, taj, top;
int head[maxn], edgenum;

struct Edge
{
	int from, to, next;
}edge[maxm];

int cnt;
vector<int>bcc[maxn];
int Tarjan(int u, int k)
{
	DFN[u] = Low[u] = ++Time;
	Stack[top++] = u;
	Instack[u] = true;
	for(int i = head[u];i != -1;i = edge[i].next)
	{
		int v = edge[i].to;
		if(v == k)	continue;
		if(DFN[v] == -1)
		{
			Tarjan(v, k);
			Low[u] = min(Low[u], Low[v]);
		}
		else if(Instack[v] && Low[u] > DFN[v])
			Low[u] = DFN[v];
	}
	if(Low[u] == DFN[u])
	{
		cnt = 0;
		taj++;
		bcc[taj].clear();
		while(1)
		{
			int now = Stack[--top];
			cnt++;
			Instack[now] = false;
			bcc[taj].push_back(now);
			if(now == u)	break;
		}
	}
	return cnt;
}

void add(int u, int v)
{
	edge[edgenum].from = u;
	edge[edgenum].to = v;
	edge[edgenum].next = head[u];
	head[u] = edgenum++;
}

void init()
{
	memset(head, -1, sizeof head);
	edgenum = 0;
}

void init_taj()
{
	memset(DFN, -1, sizeof DFN);
	memset(Instack, false, sizeof Instack);
	Time = taj = top = 0;
}

int Find_max()
{
	int maxx = -1;
	for(int i = 1;i<=taj;i++)
		maxx = max(maxx, (int)bcc[i].size());
	return maxx;
}

int main()
{
	while(~scanf("%d%d", &n, &m))
	{
		init();
		while(m--)
		{
			int u, v;
			scanf("%d%d", &u, &v);
			add(u, v);
		}
		int ans = inf;
		for(int t = 0;t<n;t++)
		{
			init_taj();
			for(int i = 0;i<n;i++)
			{
				if(i == t)	continue;
				if(DFN[i] == -1)
				Tarjan(i, t);
			}
			int tmp = Find_max();
			ans = min(ans, tmp);
		}
		if(ans == 1)	ans = 0;
		printf("%d\n", ans);		
	}
	return 0;
}