POJ 2186 Popular Cows (強聯通)

http://poj.org/problem?id=2186

Popular Cows

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 23819		Accepted: 9767

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

題意：

一羣牛中找被其他所有牛認爲是受歡迎的牛的數量，其中受歡迎有傳遞性，比如A認爲B受歡迎，B認爲C受歡迎，那麼A認爲C也是受歡迎的。

分析：

如果某頭牛是受歡迎的，那麼從其他所有牛出發都能到達這頭牛，如果用搜索做似乎太過複雜。首先進行強聯通縮點，這樣得到一個DAG，如果該DAG有且僅有一個出度爲0的縮點點（極大強聯通分量），那麼這個縮點包括的牛的數量即爲答案。

/*
 *
 * Author : fcbruce <[email protected]>
 *
 * Time : Tue 14 Oct 2014 03:00:16 PM CST
 *
 */
#include <cstdio>
#include <iostream>
#include <sstream>
#include <cstdlib>
#include <algorithm>
#include <ctime>
#include <cctype>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <list>
#include <vector>
#include <map>
#include <set>
#define sqr(x) ((x)*(x))
#define LL long long
#define itn int
#define INF 0x3f3f3f3f
#define PI 3.1415926535897932384626
#define eps 1e-10

#ifdef _WIN32
  #define lld "%I64d"
#else
  #define lld "%lld"
#endif

#define maxm 50007
#define maxn 10007

using namespace std;

int n,m;
int fir[maxn];
int u[maxm],v[maxm],nex[maxm];
int e_max;

int pre[maxn],low[maxn],sccno[maxn],w[maxn];
int st[maxn],top;
int scc_cnt,dfs_clock;

int deg[maxn];

inline void add_edge(int s,int t)
{
  int e=e_max++;
  u[e]=s;v[e]=t;
  nex[e]=fir[u[e]];fir[u[e]]=e;
}

void tarjan_dfs(int s)
{
  pre[s]=low[s]=++dfs_clock;
  st[++top]=s;
  for (int e=fir[s];~e;e=nex[e])
  {
    int t=v[e];
    if (pre[t]==0)
    {
      tarjan_dfs(t);
      low[s]=min(low[s],low[t]);
    }
    else
    {
      if (sccno[t]==0)
        low[s]=min(low[s],pre[t]);
    }
  }

  if (pre[s]==low[s])
  {
    scc_cnt++;
    for (;;)
    {
      int x=st[top--];
      sccno[x]=scc_cnt;
      w[scc_cnt]++;
      if (x==s) break;
    }
  }
}

void find_scc()
{
  top=-1;
  scc_cnt=dfs_clock=0;
  memset(pre,0,sizeof pre);
  memset(low,0,sizeof low);
  memset(w,0,sizeof w);
  for (int i=1;i<=n;i++)
    if (pre[i]==0) tarjan_dfs(i);
}

int main()
{
#ifdef FCBRUCE
  freopen("/home/fcbruce/code/t","r",stdin);
#endif // FCBRUCE

  scanf("%d%d",&n,&m);

  e_max=0;
  memset(fir,-1,sizeof fir);

  for (int e=0,u,v;e<m;e++)
  {
    scanf("%d%d",&u,&v);
    add_edge(u,v);
  }

  find_scc();

  memset(deg,0,sizeof deg);

  for (int e=0;e<e_max;e++)
  {
    if (sccno[u[e]]==sccno[v[e]]) continue;
    deg[sccno[u[e]]]++;
  }

  int cnt=0,the_one;

  for (int i=1;i<=scc_cnt;i++)
  {
    if (deg[i]==0)
    {
      cnt++;
      the_one=i;
    }
  }

  if (cnt==1) printf("%d\n",w[the_one]);
  else  puts("0");


  return 0;
}