POJ 2186 Popular Cows (强联通)

id=2186">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

field=source&key=USACO+2003+Fall" style="text-decoration:none">USACO 2003 Fall

题意：

一群牛中找被其它全部牛觉得是受欢迎的牛的数量。当中受欢迎有传递性。比方A觉得B受欢迎。B觉得C受欢迎，那么A觉得C也是受欢迎的。

分析：

假设某头牛是受欢迎的。那么从其它全部牛出发都能到达这头牛。假设用搜索做似乎太过复杂。首先进行强联通缩点。这样得到一个DAG，假设该DAG有且仅有一个出度为0的缩点点（极大强联通分量），那么这个缩点包含的牛的数量即为答案。

/*
 *
 * Author : fcbruce <fcbruce8964@gmail.com>
 *
 * 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;
}