POJ 2186：Popular Cows Tarjan模板题

Popular Cows

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 25945		Accepted: 10612

Description

Every cow's dream is to become the most popular cow in the herd. In a herd of N (1 <= N <= 10,000) cows, you are given up to M (1 <= M <= 50,000) ordered pairs of the form (A, B) that tell you that cow A thinks that cow B is popular. Since popularity is transitive,
if A thinks B is popular and B thinks C is popular, then A will also think that C is 

popular, even if this is not explicitly specified by an ordered pair in the input. Your task is to compute the number of cows that are considered popular by every other cow. 

Input

* Line 1: Two space-separated integers, N and M 



* Lines 2..1+M: Two space-separated numbers A and B, meaning that A thinks B is popular. 

Output

* Line 1: A single integer that is the number of cows who are considered popular by every other cow. 

Sample Input

3 3
1 2
2 1
2 3

Sample Output

1

Hint

Cow 3 is the only cow of high popularity.

题意是给你几头牛(......)，然后给你一个A牛喜欢B牛 这类的关系，这种喜欢的关系还可以传递，比如A喜欢B，B喜欢C，那就A也喜欢C了。问有多少头牛被所有牛喜欢。

Tarjan模板题，话说看懂了Tarjan之后还是很爽的。

用Tarjan算法缩点之后，要出度为0的点只有一个才满足要求，想象一下，要是两个的话，就不会有牛被所有其他牛喜欢的。

然后看出度为0的点内又有多少个点即可。

代码：

#include <iostream>
#include <string>
#include <cstring>
#include <queue>
#pragma warning(disable:4996)
using namespace std;  

int head[10005],LOW[10005],DFN[10005],instack[10005],Stack[10005],Belong[10005],out[10005];
int n,m,edge_num,Dindex,Stop,Bcnt;

struct edge{
	int to;
	int next;
}Edge[50005];

void init()
{
	edge_num=0;
	Stop=Bcnt=Dindex=0;

	memset(Edge,-1,sizeof(Edge));
	memset(head,-1,sizeof(head));
	memset(LOW,0,sizeof(LOW));
	memset(DFN,0,sizeof(DFN));
	memset(instack,0,sizeof(instack));
	memset(Stack,0,sizeof(Stack));
	memset(Belong,0,sizeof(Belong));
	memset(out,0,sizeof(out));
}

void addedge(int u,int v)
{
	Edge[edge_num].to=v;
	Edge[edge_num].next=head[u];
	head[u]=edge_num;
	edge_num++;
}

void tarjan(int i)
{
	int j;
	DFN[i]=LOW[i]=++Dindex;
	instack[i]=true;
	Stack[++Stop]=i;

	for(j=head[i];j!=-1;j=Edge[j].next)
	{
		int v=Edge[j].to;
		if(DFN[v]==0)
		{
			tarjan(v);
			LOW[i]=min(LOW[i],LOW[v]);
		}
		else if(instack[v]==1)
		{
			LOW[i]=min(LOW[i],DFN[v]);
		}
	}

	if(DFN[i]==LOW[i])
	{
		Bcnt++;
		do
		{
			j=Stack[Stop--];
			instack[j]=false;
			Belong[j]=Bcnt;
		}
		while(j!=i);
	}
}

void solve()
{
	int i,j,u,v;
	init();

	cin>>n>>m;
	for(i=1;i<=m;i++)
	{
		cin>>u>>v;
		addedge(u,v);
	}

	for(i=1;i<=n;i++)
	{
		if(!DFN[i])
			tarjan(i);
	}
	for(i=1;i<=n;i++)
	{
		for(j=head[i];j!=-1;j=Edge[j].next)
		{
			if(Belong[i]!=Belong[Edge[j].to])
				out[Belong[i]]++;//计算缩点后每个点的出度
		}
	}
	int out_num=0,import;
	for(i=1;i<=Bcnt;i++)
	{
		if(!out[i])
		{
			out_num++;
			import=i;
		}
	}
	int temp=0;
	if(out_num==1)
	{
		for(i=1;i<=n;i++)
		{
			if(Belong[i]==import)
			{
				temp++;
			}
		}
		cout<<temp<<endl;
	}
	else
	{
		cout<<0<<endl;
	}

}

int main()
{
	//freopen("i.txt","r",stdin);
	//freopen("o.txt","w",stdout);

	solve();
	return 0;
}  

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