poj--2186--Popular Cows (scc+缩点）

Popular Cows

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 2   Accepted Submission(s) : 1

Problem 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

/*找出最受欢迎的牛*/
#include<stdio.h>
#include<string.h>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
using namespace std;
#define MAX 50010
struct node
{
	int u,v;
	int next;
}edge[MAX];
int low[MAX],dfn[MAX];
int sum,sumin,sumout;
int sccno[MAX],scc_cnt;
int head[MAX],dfs_clock,cnt;
bool Instack[MAX];
vector<int>G[MAX];
vector<int>scc[MAX];
stack<int>s;
int in[MAX],out[MAX];
int m,n;
void init()
{
	memset(head,-1,sizeof(head));
	cnt=0;
}
void add(int u,int v)
{
	edge[cnt].u=u;
	edge[cnt].v=v;
	edge[cnt].next=head[u];
	head[u]=cnt++;
}
void getmap()
{
	int a,b;
	while(m--)
	{
		scanf("%d%d",&a,&b);
		add(a,b);
	}
}
void tarjan(int u,int fa)
{
	int v;
	low[u]=dfn[u]=++dfs_clock;
	s.push(u);
	Instack[u]=true;
	for(int i=head[u];i!=-1;i=edge[i].next)
	{
		v=edge[i].v;
		if(!dfn[v])
		{
			tarjan(v,u);
			low[u]=min(low[u],low[v]);
		}
		else if(Instack[v])
		low[u]=min(low[u],dfn[v]);
	}
	if(low[u]==dfn[u])
	{
		scc_cnt++;
		scc[scc_cnt].clear();
		for(;;)
		{
			v=s.top();
			s.pop();
			Instack[v]=false;
			scc[scc_cnt].push_back(v);
			sccno[v]=scc_cnt;
			if(u==v) break;
		}
	}
}
void find(int l,int r)
{
	memset(low,0,sizeof(low));
	memset(dfn,0,sizeof(dfn));
	memset(sccno,0,sizeof(sccno));
	memset(Instack,false,sizeof(Instack));
	scc_cnt=dfs_clock=0;
	for(int i=l;i<=r;i++)
	if(!dfn[i])
	tarjan(i,-1);
}
void suodian()
{
	for(int i=1;i<=scc_cnt;i++)
	G[i].clear(),in[i]=0,out[i]=0;
	for(int i=0;i<cnt;i++)
	{
		int u=sccno[edge[i].u];
		int v=sccno[edge[i].v];
		if(v!=u)
		{
			G[u].push_back(v);
			out[u]++,in[v]++;
		}
	}
}
void solve()
{
	int sum=0;
	int ans=0;
	for(int i=1;i<=scc_cnt;i++)
	{
		if(out[i]==0)
		//出度为零说明在新图里他是叶子节点，现在统计叶子节点个数 
		{
			sum++;
			ans+=scc[i].size();
		}
	}
	if(sum>1) printf("0
");
	if(sum==0) printf("%d
",n);
	//如果只有1个scc，说明全部的牛都是最受欢迎的 
	if(sum==1) printf("%d
",ans);
	//如果sum==0说明在新图里叶子节点只有一个
	//那么当前的scc中所有的牛都是最受欢迎的 
}
int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		init();
		getmap();
		find(1,n);
		suodian();
		solve();
	}
	return 0;
}