杭电 3836 多校联合赛 强连通分量

     一道多校联合赛的题，就是因为这道题才看强连通分量的，，，看了几天，现在算是明白了。。。具体到这道题，求出强连通分量的个数后，再统计出度为0的强连通分量的个数和入度为0的强连通分量的个数，取较大值即可。。。。。。。题目：

Equivalent Sets

Time Limit: 12000/4000 MS (Java/Others)    Memory Limit: 104857/104857 K (Java/Others)
Total Submission(s): 1155    Accepted Submission(s): 367

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. 
    
    
    
    
     
   
   
   
   

 

ac代码：

#include <iostream>
#include <string.h>
#include <cstdio>
#include <vector>
using namespace std;
int n,m,index,cnt,top;
const int N=20010;
int in[N],out[N],dfn[N],low[N];
int stack[N],belong[N];
bool instack[N];
vector<int> ss[N];
void init(){
  index=cnt=top=0;
  memset(ss,0,sizeof(ss));
  memset(dfn,0,sizeof(dfn));
  memset(low,0,sizeof(low));
  memset(belong,0,sizeof(belong));
  memset(stack,0,sizeof(stack));
  memset(instack,0,sizeof(instack));
  memset(in,0,sizeof(in));
  memset(out,0,sizeof(out));
}
int min(int x,int y){
  if(x>y)
      return y;
  else
      return x;
}
int max(int x,int y){
  if(x>y)
      return x;
  else
      return y;
}
void targin(int x){
    dfn[x]=low[x]=++index;
    instack[x]=true;
    stack[++top]=x;
    int j;
    for(int i=0;i<ss[x].size();++i){
        if(dfn[ss[x][i]]==0){
          targin(ss[x][i]);
          low[x]=min(low[ss[x][i]],low[x]);
        }
        else if(instack[ss[x][i]]){
          low[x]=min(low[x],dfn[ss[x][i]]);
        }
    }
    if(low[x]==dfn[x]){
      cnt++;
      do{
       j=stack[top--];
       belong[j]=cnt;
       instack[j]=false;
      }while(x!=j);
    }
}
int main(){
    while(~scanf("%d%d",&n,&m)){
      if(m==0)
          printf("%d\n",n);
      else{
        init();
        int x,y;
        while(m--){
          scanf("%d%d",&x,&y);
          ss[x].push_back(y);
        }
        for(int i=1;i<=n;++i){
          if(dfn[i]==0)
              targin(i);
        }
        if(cnt==1)
            printf("0\n");
        else{
              for(int i=1;i<=n;++i){
              for(int j=0;j<ss[i].size();++j){
                int x=belong[i];
                int y=belong[ss[i][j]];
                if(x!=y)
                {
                  out[x]=1;in[y]=1;
                }
              }
          }
          int xx=0,yy=0;
          for(int i=1;i<=cnt;++i){
            if(!in[i])
                xx++;
            if(!out[i])
                yy++;
          }
          printf("%d\n",max(xx,yy));
        }
      }
    }
  return 0;
}