poj Popular Cows（tarjan +缩点）

最新推荐文章于 2019-11-19 21:25:47 发布

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#tarjan #强连通

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本文探讨了在一个包含N头牛的群体中，通过M对有序对来评估每头牛的流行度，利用Tarjan算法进行点的合并和强连通分支的识别，最终确定哪些牛被所有其他牛认为是最受欢迎的。通过实例输入，展示了如何计算并理解特定牛的高受欢迎度。

Language:

Popular Cows

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 24384		Accepted: 10007

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

题意: n头牛相互羡慕，可以传递羡慕，问被所有牛羡慕的牛的最大数目

思路：

tarjan缩点，把不同的点规划到一个强联通分支，最后判断强连通分支出度为0的点是否唯一，唯一则有解，答案是该强连通分支的点的数目，否者无解

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


#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __int64 ll;

#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define bug         pf("Hi\n")

using namespace std;

#define N 10024

int head[N],low[N],time[N],e_num,tim_num;
int n,m,ans,instack[N],vis[N],type[N],out[N];

struct stud{
  int to,next;
}e[N*10];

stack<int>q;

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

void tarjan(int x)
{
    int i,j;
    q.push(x);
    instack[x]=1;

    time[x]=low[x]=++tim_num;

    for(i=head[x];i!=-1;i=e[i].next)
    {
        int to=e[i].to;
        if(time[to]==0)
        {
            tarjan(to);
            if(low[x]>low[to]) low[x]=low[to];
        }
        else
            if(instack[to]&&low[x]>time[to])
                low[x]=time[to];
    }

    if(low[x]==time[x])
    {
        ans++;
        do{
            j=q.top();
            q.pop();
            instack[j]=0;
            type[j]=ans;   //标记j 属于第几个联通分支
            vis[ans]++;    //这个联通分支的点+1
           // printf("%d ans=%d\n",j,ans);
        }while(j!=x);
    }
}

void solve()
{
    int i,j;
    mem(time,0);
    mem(instack,0);
    mem(out,0);
    mem(vis,0);

    ans=tim_num=0;

    for(i=1;i<=n;i++)
        if(time[i]==0)
          tarjan(i);

	for(i=1;i<=n;i++)
		for(j=head[i];j!=-1;j=e[j].next)
	{
		int to=e[j].to;

		if(type[i]!=type[to])
			{
				out[type[i]]=1;  //标记有出度的联通分支
			}
	}

	int t=0,pos;

	for(i=1;i<=ans;i++)    //一共有ans个联通分支，取出其中
		if(out[i]==0)
	{
		t++;
		pos=i;
		if(t>1) break;
	}

   //如果出度为0的联通分支只有一个，那么这个联通分支的个数就是答案,
   //否则，无解
	if(t==1)
		printf("%d\n",vis[pos]);
	else
		printf("0\n");


}

int main()
{
    int i,j;
    scanf("%d%d",&n,&m);
    {
        int u,v;
        e_num=0;

        mem(head,-1);

        while(m--)
        {
            sff(u,v);
            add(u,v);
        }
        solve();
    }
    return 0;
}