poj2186 Popular Cows(强连通缩点)

最新推荐文章于 2020-05-18 14:45:19 发布

原创最新推荐文章于 2020-05-18 14:45:19 发布 · 381 阅读

0 ·

CC 4.0 BY-SA版权

强连通专栏收录该内容

7 篇文章

订阅专栏

本文介绍了一种算法，用于解决牛群中流行的牛只数量计算问题。通过输入成对的牛只关系，该算法能够找出被所有其他牛只认为是流行的牛只数量。涉及到图论中的连通性和传递闭包概念。

Popular Cows

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 28016		Accepted: 11294

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

Sample Output

Hint

Cow 3 is the only cow of high popularity.

Source

USACO 2003 Fall

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<stack>
#include<queue>
#include<cmath>
#include<map>
#include<set>

typedef long long ll;
#define eps 1e-8

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)
#define bug printf("hi\n")
using namespace std;
#define INF 0x3f3f3f3f
#define N 10005

int head[N],dfn[N],low[N],belong[N];
bool instack[N];
int all,Time,num;
int out[N];
int n,m;
stack<int>s;

struct stud{
   int to,ne;
}e[N*5];

inline void add(int u,int v)
{
	e[num].to=v;
	e[num].ne=head[u];
	head[u]=num++;
}

void dfs(int u,int fa)
{
//	printf("%d\n",u);
    dfn[u]=low[u]=++Time;
	s.push(u);
	instack[u]=true;
	for(int i=head[u];i!=-1;i=e[i].ne)
	{
		int to=e[i].to;
	    if(dfn[to]==-1)
		{
			dfs(to,i);
			low[u]=min(low[u],low[to]);
		}
		else if(instack[to])
			low[u]=min(low[u],low[to]);
	}
	if(low[u]==dfn[u])
	{
		++all;
		int temp;
		do{
            temp=s.top();
		   s.pop();
		   instack[temp]=false;
		   belong[temp]=all;
		}while(temp!=u);
	}
}

int main()
{
	int i,j;
    while(~scanf("%d%d",&n,&m))
	{
		memset(head,-1,sizeof(head));
		num=0;
		int u,v;
		while(m--)
		{
			scanf("%d%d",&u,&v);
			add(u,v);
		}
		memset(dfn,-1,sizeof(dfn));
		memset(low,-1,sizeof(low));
		memset(instack,false,sizeof(instack));
        while(!s.empty()) s.pop();
		all=0;
	    Time=0;
  //  bug;
		for(i=1;i<=n;i++)
			if(dfn[i]==-1) 
				dfs(i,-1);
	//	bug;
       memset(out,0,sizeof(out));
	   for(i=1;i<=n;i++)
		   for(j=head[i];j!=-1;j=e[j].ne)
		   {
			   int to=e[j].to;
			   if(belong[i]==belong[to]) continue;
			   out[belong[i]]++;
		   }
	   int cnt=0,last;
	   for(i=1;i<=all;i++)
		   if(out[i]==0)
		   {
			   cnt++;
			   last=i;
		   }
	   if(cnt!=1)
	   {
		   printf("0\n");
		   continue;
	   }
	   int ans=0;
	   for(i=1;i<=n;i++)
		   if(belong[i]==last) ans++;
	   printf("%d\n",ans);
	}
	return 0;
}