hdu3836联通的强还原性点

Equivalent Sets

Time Limit: 12000/4000 MS (Java/Others)    Memory Limit: 104857/104857 K (Java/Others)
Total Submission(s): 2526    Accepted Submission(s): 857

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

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include <map>
#include <cmath>
#include <iomanip>
#define INF 99999999
typedef long long LL;
using namespace std;

const int MAX=20000+10;
int n,m,size,top,index,ind,oud;
int head[MAX],dfn[MAX],low[MAX],stack[MAX];
int mark[MAX],flag[MAX];
//dfn表示点u出现的时间,low表示点u能到达所属环中最早出现的点(记录的是到达的时间) 

struct Edge{
	int v,next;
	Edge(){}
	Edge(int V,int NEXT):v(V),next(NEXT){}
}edge[50000+10];

void Init(int num){
	for(int i=0;i<=num;++i)head[i]=-1;
	size=top=index=ind=oud=0;
}

void InsertEdge(int u,int v){
	edge[size]=Edge(v,head[u]);
	head[u]=size++;
}

void tarjan(int u){
	if(mark[u])return;
	dfn[u]=low[u]=++index;
	stack[++top]=u;
	mark[u]=1;
	for(int i=head[u];i != -1;i=edge[i].next){
		int v=edge[i].v;
		tarjan(v);
		if(mark[v] == 1)low[u]=min(low[u],low[v]);//必须点v在栈里面才行 
	}
	if(dfn[u] == low[u]){
		++ind,++oud;//记录缩点之后的点的个数,方便计算入度和出度为0的点的个数
		while(stack[top] != u){
			mark[stack[top]]=-1;
			low[stack[top--]]=low[u];
		}
		mark[u]=-1;
		--top;
	}
}

int main(){
	int u,v;
	while(~scanf("%d%d",&n,&m)){
		Init(n);
		for(int i=0;i<m;++i){
			scanf("%d%d",&u,&v);
			InsertEdge(u,v);
		}
		memset(mark,0,sizeof mark);
		for(int i=1;i<=n;++i){
			if(mark[i])continue;
			tarjan(i);
		}
		if(ind == 1){printf("0
");continue;}//仅仅剩一个点了表示原图数个强联通图 
		for(int i=0;i<=n;++i)mark[i]=flag[i]=0;
		for(int i=1;i<=n;++i){
			for(int j=head[i];j != -1;j=edge[j].next){
				v=edge[j].v;
				if(low[i] == low[v])continue;
				if(mark[low[i]] == 0)--oud;//mark标记点i是否已有出度 
				if(flag[low[v]] == 0)--ind;//flag标记点v是否已有入度
				mark[low[i]]=flag[low[v]]=1;
			}
		}
		printf("%d
",max(ind,oud));
	} 
	return 0;
}

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