poj2186 Popular Cows 有向图 强连通分量

Popular Cows

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 20718 Accepted: 8438

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 31 22 12 3

Sample Output

1

Hint

Cow 3 is the only cow of high popularity.

Source

USACO 2003 Fall

给个有向图，先求强连通分量,缩点之后,如果出度为0的只有一个,那个结果自然是这个缩点的点数,如果为0的大于1,则一定不连通,所以直接输出0就可以了!

#include <iostream>#include <stdio.h>#include <string.h>#include <stdlib.h>#define N 10510#define E 100000#define M 1000000using namespace std;int head[N],sta[M],num[N],re,ans,next[E],id[N],in[N],vec[E],vis[N],dfn[N],low[N],clock_m,edge_m;int addedge(int s,int e){    vec[edge_m]=e;next[edge_m]=head[s];head[s]=edge_m++;}int init(){    memset(vis,0,sizeof(vis));    memset(dfn,0,sizeof(dfn));    memset(low,0,sizeof(low));    memset(num,0,sizeof(num));    memset(head,-1,sizeof(head));    memset(id,0,sizeof(id));    memset(in,0,sizeof(in));    edge_m=0;clock_m=0;ans=0;re=0;}int tarjan(int x){    dfn[x]=low[x]=++clock_m;    sta[++ans]=x;    vis[x]=1;    for(int i=head[x];i!=-1;i=next[i]){        int goal=vec[i];        if(!dfn[goal]){            tarjan(goal);            low[x]=min(low[x],low[goal]);        }        else if(/*vis[goal]*/!id[goal])            low[x]=min(low[x],dfn[goal]);    }    if(low[x]==dfn[x]){        re++;int v;       do{            v=sta[ans--];vis[v]=0;num[re]++;id[v]=re;        }while(v!=x);    }    return 1;}int main(){    int tcase,n,m,s,e;    while(scanf("%d%d",&n,&m)!=EOF)    {        //system("PAUSE");        init();        for(int i=0;i<m;i++){            scanf("%d%d",&s,&e);            addedge(s,e);        }        for(int i=1;i<=n;i++)        if(!dfn[i])        {           tarjan(i);        }        for(int i=1;i<=n;i++){            for(int j=head[i];j!=-1;j=next[j]){                if(id[vec[j]]!=id[i]){                    in[id[i]]++;                }            }        }        ans=0;int k=1;        for(int i=1;i<=re;i++){            if(in[i]==0){            sta[ans++]=i;k=i;            }        }        if(ans>1)printf("0\n");        else printf("%d\n",num[k]);    }    return 0;}