Equivalent Sets

Time Limit: 12000/4000 MS (Java/Others)    Memory Limit: 104857/104857 K (Java/Others)

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.

Source

2011 Multi-University Training Contest 1 - Host by HNU

题意：给你n个点，m条边的有向图，最少加几条边使得改图为强连通；

思路：对于一个缩完点的图，要使得其强连通，入度和出度都至少为1；

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
#include<stdlib.h>
#include<time.h>
using namespace std;
#define LL long long
#define pi (4*atan(1.0))
#define eps 1e-6
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=1e5+10,M=1e6+10,inf=1e9+10;
const LL INF=5e17+10,mod=1e9+7;

struct is
{
    int u,v;
    int next;
}edge[50010];
int head[50010];
int belong[50010];
int dfn[50010];
int low[50010];
int stackk[50010];
int instack[50010];
int number[50010];
int in[N],out[N];
int n,m,jiedge,lu,bel,top;
void update(int u,int v)
{
    jiedge++;
    edge[jiedge].u=u;
    edge[jiedge].v=v;
    edge[jiedge].next=head[u];
    head[u]=jiedge;
}
void dfs(int x)
{
    dfn[x]=low[x]=++lu;
    stackk[++top]=x;
    instack[x]=1;
    for(int i=head[x];i;i=edge[i].next)
    {
        if(!dfn[edge[i].v])
        {
            dfs(edge[i].v);
            low[x]=min(low[x],low[edge[i].v]);
        }
        else if(instack[edge[i].v])
        low[x]=min(low[x],dfn[edge[i].v]);
    }
    if(low[x]==dfn[x])
    {
        int sum=0;
        bel++;
        int ne;
        do
        {
            sum++;
            ne=stackk[top--];
            belong[ne]=bel;
            instack[ne]=0;
        }while(x!=ne);
        number[bel]=sum;
    }
}
void tarjan()
{
    memset(dfn,0,sizeof(dfn));
    bel=lu=top=0;
    for(int i=1;i<=n;i++)
    if(!dfn[i])
    dfs(i);
}
int main()
{
    int i,t;
    while(~scanf("%d%d",&n,&m))
    {
        memset(in,0,sizeof(in));
        memset(out,0,sizeof(out));
        memset(head,0,sizeof(head));
        jiedge=0;
        for(i=1;i<=m;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            update(u,v);
        }
        tarjan();
        int x=0;
        int z=0;
        for(i=1;i<=jiedge;i++)
        if(belong[edge[i].v]!=belong[edge[i].u])
        {
            if(!out[belong[edge[i].u]])x++;
            if(!in[belong[edge[i].v]])z++;
            out[belong[edge[i].u]]++;
            in[belong[edge[i].v]]++;
        }
        x=bel-x;
        z=bel-z;
        if(bel==1)
            printf("0\n");
        else
            printf("%d\n",max(x,z));
    }
    return 0;
}