Equivalent Sets

Time Limit: 12000/4000 MS (Java/Others)    Memory Limit: 104857/104857 K (Java/Others)
Total Submission(s): 3568    Accepted Submission(s): 1235

Problem Description

To prove two sets A and B are equivalent, we can first prove A is a subset of B, and then prove B is a subset of A, so finally we got that these two sets are equivalent.
You are to prove N sets are equivalent, using the method above: in each step you can prove a set X is a subset of another set Y, and there are also some sets that are already proven to be subsets of some other sets.
Now you want to know the minimum steps needed to get the problem proved.

Input

The input file contains multiple test cases, in each case, the first line contains two integers N <= 20000 and M <= 50000.
Next M lines, each line contains two integers X, Y, means set X in a subset of set Y.

Output

For each case, output a single integer: the minimum steps needed.

Sample Input

4 0

3 2

1 2

1 3

Sample Output

4

2

Hint

Case 2: First prove set 2 is a subset of set 1 and then prove set 3 is a subset of set 1.

题意：n个点m条边的有向图，问最少增加多少边使图强连通。

题解：求每个scc的入度和出度，然后分别求出入度中0的个数in和出度out，取in和out中较大的一个；

因为入度或出度为0证明这个scc和别的scc未相连，需要用一条边相连，这条边就是要加入的边，又因为一个scc可能连接多个scc，即只考虑入度或者只考虑出度都不准确

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stack>
#include<vector>
#define MAX 50010
#define INF 0x3f3f3f
using namespace std;
int n,m;
int ans,head[MAX];
int low[MAX],dfn[MAX];
int instack[MAX],sccno[MAX];
vector<int>newmap[MAX];
vector<int>scc[MAX];
int scccnt,dfsclock;
int in[MAX],out[MAX];
stack<int>s;
struct node
{
	int beg,end,next;
}edge[MAX];
void init()
{
	ans=0;
	memset(head,-1,sizeof(head));
}
void add(int beg,int end)
{
	edge[ans].beg=beg;
	edge[ans].end=end;
	edge[ans].next=head[beg];
	head[beg]=ans++;
}
void getmap()
{
	int i,a,b;
	while(m--)
	{
		scanf("%d%d",&a,&b);
		add(a,b);
	}
}
void tarjan(int u)
{
	int v,i,j;
	s.push(u);
	instack[u]=1;
	low[u]=dfn[u]=++dfsclock;
	for(i=head[u];i!=-1;i=edge[i].next)
	{
		v=edge[i].end;
		if(!dfn[v])
		{
			tarjan(v);
			low[u]=min(low[u],low[v]);
		}
		else if(instack[v])
		low[u]=min(low[u],dfn[v]);
	}
	if(low[u]==dfn[u])
	{
	    scccnt++;
		while(1)
		{
			v=s.top();
			s.pop();
			instack[v]=0;
			sccno[v]=scccnt;
			if(v==u)
			break;
		}
	}
}
void find(int l,int r)
{
	memset(low,0,sizeof(low));
	memset(dfn,0,sizeof(dfn));
	memset(instack,0,sizeof(instack));
	memset(sccno,0,sizeof(sccno));
	dfsclock=scccnt=0;
	for(int i=l;i<=r;i++)
	{
		if(!dfn[i])
		    tarjan(i);
	}
}
void suodian()
{
	int i;
	for(i=1;i<=scccnt;i++)
	{
		newmap[i].clear();
		in[i]=0;out[i]=0;
	}
	for(i=0;i<ans;i++)
	{
		int u=sccno[edge[i].beg];
		int v=sccno[edge[i].end];
		if(u!=v)
		{
			newmap[u].push_back(v);
			in[v]++;out[u]++;
		}
	}
}
void solve()
{
	int i,j;
	if(scccnt==1)
	{
		printf("0\n");
		return ;
	}
	else
	{
		int minn=0;
		int maxx=0;
		for(i=1;i<=scccnt;i++)
		{
			if(!in[i])
			    minn++;
			if(!out[i])
			    maxx++;
		}
		printf("%d\n",max(minn,maxx));
	}
}
int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		init();
		getmap();
		find(1,n);
		suodian();
		solve();
	}
	return 0;
}