poj 2186 Popular Cows

题目：

Popular Cows

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 18659		Accepted: 7516

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.

代码：

#include<iostream>
#include<stdio.h>
#include<stack>
#include<string.h>
using namespace std;
#define MAXN 10010
struct Node
{
int v;
int next;
} edge[50010];
int head[MAXN];
int dfn[MAXN];
int low[MAXN];
bool mark[MAXN]; //to judge in stack or not
int id[MAXN]; //the id of scc
int T;
stack<int> sta;
int scc;
void tarjan(int v)
{
dfn[v] = low[v] = ++T;
sta.push(v);
mark[v] = true;
int k;
for (k = head[v]; k != -1; k = edge[k].next)
{
int u = edge[k].v;
if (!dfn[u])
{
tarjan(u);
low[v] = min(low[v], low[u]);
}
else if (mark[u])
low[v] = min(low[v], dfn[u]);
}
if (low[v] == dfn[v])
{
++scc;
int u;
do
{
u = sta.top();
sta.pop();
mark[u] = false;
id[u] = scc;
}
while (u != v);
}
return;
}
void solve(int n)
{
scc = 0;
int i;
for (i = 1; i <= n; ++i)
if (!dfn[i])
tarjan(i);
}
int main()
{
int n,m;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
{
head[i]=-1;
mark[i]=false;
dfn[i]=0;
}
scc=0;
int a,b;
for(int i=0;i<m;i++)
{
scanf("%d%d",&a,&b);
edge[i].v=b;
edge[i].next=head[a];
head[a]=i;
}
solve(n);

memset(mark,false,sizeof(mark));
for(int i=1;i<=n;i++)
{
for(int j=head[i];j!=-1;j=edge[j].next)
{
if(id[i]!=id[edge[j].v])
{
mark[id[i]]=true;
}
}
}
int ans=0;
int idd;
for(int i=1;i<=scc;i++)
{
if(!mark[i])
{
ans++;
idd=i;
}
}
if(ans>1)
{
printf("0\n");
}
else
{
ans=0;
for(int i=1;i<=n;i++)
{
if(id[i]==idd)
ans++;
}
printf("%d\n",ans);
}
return 0;
}