poj 2186 Popular Cows

Popular Cows

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 29908		Accepted: 12131

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

题解

tarjan一直没学会，今天写了道入门题。主要思路是用tarjan()求出强连通分量，记录每一个强连通分量的出度，出度为0的就是大家都认可的popular cow。在数组问题上卡了很久，后来发现u[i]，v[i]要开5*10100才ac。

15701878

ksq2013

2186

Accepted

1304K

94MS

C++

1969B

2016-07-11 13:36:28

#include<cstdio>
#include<cstring>
#include<iostream>
#define M 10100
using namespace std;
int n,m,u[M*],v[M*],first[M*],mnext[M*],ans[M],outdu[M];
int dfs_clock,scc_cnt,stack_count,my_stack[M*],pre[M],sccno[M],lowlink[M];
void Init()
{
    memset(first,-,sizeof(first));
    memset(mnext,,sizeof(mnext));
    memset(my_stack,,sizeof(my_stack));
    memset(lowlink,,sizeof(lowlink));
    memset(sccno,,sizeof(sccno));
    memset(pre,,sizeof(pre));
    memset(ans,,sizeof(ans));
    memset(outdu,,sizeof(outdu));
}
void Link()
{
    for(int i=;i<=m;i++){
        scanf("%d%d",&u[i],&v[i]);
        mnext[i]=first[u[i]];
        first[u[i]]=i;
    }
}
void dfs(int ux)
{
    pre[ux]=lowlink[ux]=++dfs_clock;
    my_stack[++stack_count]=ux;
    int vx;
    for(int i=first[ux];i!=-;i=mnext[i]){
        vx=v[i];
        if(!pre[vx]){
            dfs(vx);
            lowlink[ux]=min(lowlink[ux],lowlink[vx]);
        }
        else if(!sccno[vx])
                 lowlink[ux]=min(lowlink[ux],pre[vx]);
    }
    if(lowlink[ux]==pre[ux]){
        scc_cnt++;
        int tmp=;
        do{
            tmp++;
            vx=my_stack[stack_count--];
            sccno[vx]=scc_cnt;
        }while(vx!=ux);
        ans[scc_cnt]=tmp;
    }
}
void Tarjan()
{
    dfs_clock=scc_cnt=stack_count=;
    for(int i=;i<=n;i++)
        if(!pre[i])dfs(i);
}
int main()
{
    while(~scanf("%d%d",&n,&m)){
        Init();
        Link();
        Tarjan();
        if(scc_cnt==){
            printf("%d\n",n);
            continue;
        }
        for(int i=;i<=n;i++){
            for(int j=first[i];j!=-;j=mnext[j]){
                int vx=v[j];
                if(sccno[i]!=sccno[vx])
                    outdu[sccno[i]]++;
            }
        }
        int res=;
        for(int i=;i<=scc_cnt;i++)
            if(!outdu[i]){
                if(!res)res=ans[i];
                else {res=;break;}
            }
        printf("%d\n",res);
    }
    return ;
}