HDU 3836 Equivalent Sets（Tarjan+縮點)

HDU 3836 Equivalent Sets（Tarjan+縮點)

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. 

強聯通縮點：添加幾條邊成強聯通分量：設縮點後所有點中出度為0的點為d_1,入度為0點為d_2，則答案為max(d_1,d_2)；

#include
#include
#include
#include
#include
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