HDOJ 题目3836 Equivalent Sets（强连通分量）

最新推荐文章于 2018-08-29 14:45:30 发布

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最新推荐文章于 2018-08-29 14:45:30 发布

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分类专栏：强连通

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本文探讨了通过证明集合间的子集关系来确定多个集合等价性的算法问题。介绍了如何利用已知的子集关系并通过添加最少的额外关系使所有集合成为强连通分量的方法。

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Equivalent Sets

Time Limit: 12000/4000 MS (Java/Others) Memory Limit: 104857/104857 K (Java/Others)
Total Submission(s): 3070 Accepted Submission(s): 1079

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

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.

Source

2011 Multi-University Training Contest 1 - Host by HNU

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xubiao | We have carefully selected several similar problems for you: 3835 3828 3834 3830 3833

题目大意：问加几个边，让他成为一个强连通分量

ac代码

#include<stdio.h>
#include<string.h>
#include<queue>
#include<string>
#include<iostream>
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
using namespace std;
int inde[50020],outde[50020];
int dfn[50020],low[50020],ins[50020],belong[50020],stack[50020],cnt,head[50050],top,taj,time;
struct s
{
	int u,v,next;
}edge[50050];
void add(int u,int v)
{
	edge[cnt].u=u;
	edge[cnt].v=v;
	edge[cnt].next=head[u];
	head[u]=cnt++;
}
void init()
{
	cnt=0;
	memset(head,-1,sizeof(head));
	top=taj=0;
	memset(dfn,-1,sizeof(dfn));
	memset(belong,-1,sizeof(belong));
	memset(low,-1,sizeof(low));
	memset(ins,0,sizeof(ins));
	time=0;
}
void tarjan(int u)
{
	dfn[u]=low[u]=time++;
	ins[u]=1;
	stack[top++]=u;
	for(int i=head[u];i!=-1;i=edge[i].next)
	{
		int v=edge[i].v;
		if(dfn[v]==-1)
		{
			tarjan(v);
			low[u]=min(low[u],low[v]);
		}
		else
			if(ins[v])
			{
				low[u]=min(low[u],dfn[v]);
			}
	}
	if(dfn[u]==low[u])
	{
		int now;
		taj++;
		do
		{
			now=stack[--top];
			ins[now]=0;
			belong[now]=taj;
		}while(now!=u);
	}
}
int n,m;
int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		int i;
		init();
		for(i=0;i<m;i++)
		{
			int a,b;
			scanf("%d%d",&a,&b);
			add(a,b);
		}
		for(i=1;i<=n;i++)
			if(dfn[i]==-1)
				tarjan(i);
		if(n<1||taj==1)
		{
			printf("0\n");
			continue;
		}
		int ans1,ans2;
		ans1=ans2=0;
		memset(inde,0,sizeof(inde));
		memset(outde,0,sizeof(outde));
		for(i=0;i<cnt;i++)
		{
			int u=edge[i].u;
			int v=edge[i].v;
			if(belong[u]!=belong[v])
			{
				inde[belong[v]]++;
				outde[belong[u]]++;
			}
		}
		for(i=1;i<=taj;i++)
		{
			if(inde[i]==0)
				ans1++;
			if(outde[i]==0)
				ans2++;
		}
		printf("%d\n",max(ans1,ans2));
	}
}