poj2086——Popular Cows（强连通分量）

Popular Cows

Time Limit: 2000MS Memory Limit: 65536K

Total Submissions: 18349Accepted: 7390

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
Hint

Cow 3 is the only cow of high popularity. 

Source

USACO 2003 Fall

解析：

       根据题意建立有向无环图，用tarjan求强连通分量，缩点。。。

       再重新建图，维护出度，找出度为零的点的个数。。。

       本题是一道较裸的求强连通分量的题。。。

代码：

#include<cstdio>#include<iostream>#include<algorithm>using namespace std;const int N=10010;const int M=50010;int l=0;struct status{    int to,next;}node1[M],node2[M];int head[N],first[N];int low[N],dfsnum[N],stack[N];int n,m,stamp,cnt,top,e=0;bool vis[N];int cnt_size[N],belong[N],outdegree[N];void add1(int u,int v){    node1[l].to=v;    node1[l].next=head[u];    head[u]=l++;}void tarjan(int u){    dfsnum[u]=low[u]=++stamp;    vis[u]=true;    stack[top++]=u;    for(int i=head[u];i!=-1;i=node1[i].next)    {        int v=node1[i].to;        if(!dfsnum[v])        {            tarjan(v);            low[u]=min(low[u],low[v]);        }        else if(vis[v])low[u]=min(dfsnum[v],low[u]);    }    if(dfsnum[u]==low[u])    {        do        {            top--;            vis[stack[top]]=false;            cnt_size[cnt]++;            belong[stack[top]]=cnt;        } while(stack[top]!=u);        cnt++;    }}void add2(int u,int v){    node2[e].to=v;    node2[e].next=first[u];    first[u]=e++;}void work(){    freopen("xx.in","r",stdin);    freopen("xx.out","w",stdout);    scanf("%d%d",&n,&m); l=0;    memset(head,-1,sizeof(head));    for(int i=1;i<=m;i++)    {        int u,v;        scanf("%d%d",&u,&v);        add1(u,v);                   //邻接表建图    }    memset(dfsnum,0,sizeof(dfsnum));memset(vis,0,sizeof(vis));stamp=cnt=top=0;memset(cnt_size,0,sizeof(cnt_size));memset(low,0,sizeof(low));    for(int i=1;i<=n;i++)    {        if(!dfsnum[i])tarjan(i);    //tarjan找强连通分量，缩点    }    memset(first,-1,sizeof(first));    for(int i=1;i<=n;i++)        for(int j=head[i];j!=-1;j=node1[j].next)        {            int v=node1[j].to;            if(belong[i]!=belong[v])            {                add2(belong[i],belong[v]);   //重新建图            }        }    memset(outdegree,0,sizeof(outdegree));    for(int i=0;i<cnt;i++)    {        for(int j=first[i];j!=-1;j=node2[j].next) outdegree[i]++;  //维护点的出度    }    int num=0;    int sum=0;    for(int i=0;i<cnt;i++)    {        if(outdegree[i]==0)      //找出度为零的节点个数        {            sum=cnt_size[i];            num++;        }    }    if(num==1)printf("%d\n",sum); else printf("0\n");    }int main(){    work();    return 0;}