HDU 3836 Equivalent SetsTarjan+缩点)

Problem Description

To prove two sets A and B are equivalent, we can first prove A is a subset of B, and then prove B is a subset of A, so finally we got that these two sets are equivalent.
You are to prove N sets are equivalent, using the method above: in each step you can prove a set X is a subset of another set Y, and there are also some sets that are already proven to be subsets of some other sets.
Now you want to know the minimum steps needed to get the problem proved.

 

Input

The input file contains multiple test cases, in each case, the first line contains two integers N <= 20000 and M <= 50000.
Next M lines, each line contains two integers X, Y, means set X in a subset of set Y.

 

Output

For each case, output a single integer: the minimum steps needed.

 

Sample Input

   
     

      
    4 0
3 2
1 2
1 3
   
     

 

Sample Output

   
     

      
    4
2

    
      
 
       
 
        Hint  
       
Case 2: First prove set 2 is a subset of set 1 and then prove set 3 is a subset of set 1. 
      
    
       
   
     

 

强联通缩点：加入几条边成强联通分量：设缩点后全部点中出度为0的点为d_1,入度为0点为d_2，则答案为max(d_1,d_2)；

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
typedef long long LL;
using namespace std;

#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )

const int maxn=20000+100;
const int maxm=100000;
struct node{
    int u,v;
    int next;
}e[maxm];
int head[maxn],cntE;
int DFN[maxn],low[maxn];
int s[maxm],top,index,cnt;
int belong[maxn],instack[maxn];
int in[maxn],out[maxn];
int n,m;
void init()
{
    top=cntE=0;
    index=cnt=0;
    CLEAR(DFN,0);
    CLEAR(head,-1);
    CLEAR(instack,0);
//    CLEAR(belong,0);
}
void addedge(int u,int v)
{
    e[cntE].u=u;e[cntE].v=v;
    e[cntE].next=head[u];
    head[u]=cntE++;
}
void Tarjan(int u)
{
    DFN[u]=low[u]=++index;
    instack[u]=1;
    s[top++]=u;
    for(int i=head[u];i!=-1;i=e[i].next)
    {
        int v=e[i].v;
        if(!DFN[v])
        {
            Tarjan(v);
            low[u]=min(low[u],low[v]);
        }
        else if(instack[v])
            low[u]=min(low[u],DFN[v]);
    }
    int v;
    if(DFN[u]==low[u])
    {
        cnt++;
        do{
            v=s[--top];
            belong[v]=cnt;
            instack[v]=0;
        }while(u!=v);
    }
}
void work()
{
    REPF(i,1,n)
      if(!DFN[i])  Tarjan(i);
    if(cnt<=1)
    {
        puts("0");
        return ;
    }
    CLEAR(in,0);
    CLEAR(out,0);
    for(int i=0;i<cntE;i++)
    {
        int u=e[i].u,v=e[i].v;
        if(belong[u]!=belong[v])
            in[belong[v]]++,out[belong[u]]++;
    }
    int d_1=0,d_2=0;
    REPF(i,1,cnt)
    {
        if(!in[i])
            d_1++;
        if(!out[i])
            d_2++;
    }
    printf("%d\n",max(d_1,d_2));
}
int main()
{
    int u,v;
    while(~scanf("%d%d",&n,&m))
    {
        init();
        for(int i=0;i<m;i++)
        {
            scanf("%d%d",&u,&v);
            addedge(u,v);
        }
        work();
    }
    return 0;
}