poj 2186 Popular Cows 【强连通分量】

Popular Cows Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 32368   Accepted: 13202 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

运用强连通分量将图变成DAG图，并知道了最后一个分量可能为拓扑终点，dfs反边搜索看是否达到所有边。

代码：

#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
int n,m,cmp[10100];
vector<int> V[10100],FV[10100],vs;
bool used[10100];
void add_edge(int a,int b)
{
V[a].push_back(b);
FV[b].push_back(a);
}
void dfs(int v)
{
used[v]=true;
for (int i=0;i<V[v].size();i++)
{
if (!used[V[v][i]]) dfs(V[v][i]);
}
vs.push_back(v);
}
void rdfs(int v,int k)
{
used[v]=true;
cmp[v]=k;
for (int i=0;i<FV[v].size();i++)
{
if (!used[FV[v][i]]) rdfs(FV[v][i],k);
}
}
int main()
{
scanf("%d%d",&n,&m);
// while (~scanf("%d%d",&n,&m))
// {
int a,b;
// for (int i=1;i<=n;i++)
// {
// V[i].clear();
// FV[i].clear();
// }
vs.clear();
for (int i=0;i<m;i++)
{
scanf("%d%d",&a,&b);
add_edge(a,b);
}
memset(used,0,sizeof(used));
for (int i=1;i<=n;i++)
{
if(!used[i]) dfs(i);
}
memset(used,0,sizeof(used));
int k=1;
for (int i=vs.size()-1;i>=0;i--)
{
if(!used[vs[i]]) rdfs(vs[i],k++);
}
int ans=0;
for (int i=1;i<=n;i++)
{
if (cmp[i]==k-1)
{
m=i;
ans++;
}
}
memset(used,0,sizeof(used));
rdfs(m,0);
for (int i=1;i<=n;i++)
{
if (!used[i])
{
ans=0;
break;
}
}
printf("%d\n",ans);
//}
return 0;
}