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

題意：給你一個有向圖，問在圖中最少要加多少條邊能使得該圖變成一個強連通圖

思路：縮點模板題

Source Program

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<string>
#include<cstdlib>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<ctime>
#include<vector>
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define N 20001
#define MOD 16007
#define E 1e-6
#define LL long long
using namespace std;
int n,m;
vector<int> G[N];
stack<int> S;
int dfn[N],low[N];
bool vis[N];
int sccno[N];
bool in[N],out[N];//記錄入度、出度是否為0
int block_cnt;
int sig;
void Tarjan(int x){
    vis[x]=true;
    dfn[x]=low[x]=++block_cnt;
    S.push(x);

    for(int i=0;i<G[x].size();i++){
        int y=G[x][i];
        if(vis[y]==false){
            Tarjan(y);
            low[x]=min(low[x],low[y]);
        }
        else if(!sccno[y])
            low[x]=min(low[x],dfn[y]);
    }

    if(dfn[x]==low[x]){
        sig++;
        while(true){
            int temp=S.top();
            S.pop();
            sccno[temp]=sig;
            if(temp==x)
                break;
        }
    }
}
int shrink(){//縮點

    memset(in,false,sizeof(in));
    memset(out,false,sizeof(out));
    for(int i=1;i<=sig;i++){
        in[i]=true;
        out[i]=true;
    }

    for(int x=0;x<n;x++){
        for(int i=0;i<G[x].size();i++){
            int y=G[x][i];
            if(sccno[x]!=sccno[y]){//統計每個點出度、入度是否為0
                out[sccno[x]]=false;
                in[sccno[y]]=false;
            }
        }
    }
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF&&(n+m))
    {
        for(int i=0;i<n;i++)
            G[i].clear();
        while(m--){
            int x,y;
            scanf("%d%d",&x,&y);
            x--;
            y--;
            G[x].push_back(y);
        }

        sig=0;
        block_cnt=0;
        memset(vis,false,sizeof(vis));
        memset(dfn,0,sizeof(dfn));
        memset(low,0,sizeof(low));
        memset(sccno,0,sizeof(sccno));

        for(int i=0;i<n;i++)
            if(vis[i]==false)
                Tarjan(i);

        shrink();
        int a=0,b=0;
        for(int i=1;i<=sig;i++){//統計入度、出度為0的點的個數
            if(in[i])
                a++;
            if(out[i])
                b++;
        }
        int res=max(a,b);
        if(sig==1)//強連通分量為1時
            res=0;

        printf("%d\n",res);
    }
}