HDU 3836 Equivalent Sets

Posted 2020-10-03 一蓑烟雨任生平

Equivalent Sets

Time Limit: 12000/4000 MS (Java/Others)    Memory Limit: 104857/104857 K (Java/Others)
Total Submission(s): 4819    Accepted Submission(s): 1733

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

 

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#include<iostream>
#include<cstdio> 
#include<cstring>
#include<algorithm>
#define MAXN 100000+15
using namespace std;
int n,m,tot,tim,top,sumcol,ans,bns;
int to[MAXN],from[MAXN],net[MAXN],x[MAXN],col[MAXN],into[MAXN],out[MAXN];
int dfn[MAXN],low[MAXN],vis[MAXN],stack[MAXN],visstack[MAXN];
void add(int u,int v){
    to[++tot]=v;x[tot]=u;net[tot]=from[u];from[u]=tot;
}
void tarjin(int now){
    low[now]=dfn[now]=++tim;
    stack[++top]=now;
    visstack[now]=1;
    vis[now]=1;
    for(int i=from[now];i;i=net[i])
        if(visstack[to[i]])
            low[now]=min(low[now],dfn[to[i]]);
        else if(!vis[to[i]]){
            tarjin(to[i]);
            low[now]=min(low[now],low[to[i]]);
        }
    if(dfn[now]==low[now]){
        sumcol++;
        col[now]=sumcol;
        while(stack[top]!=now){
            col[stack[top]]=sumcol;
            visstack[stack[top]]=0;
            top--;
        }
        visstack[now]=0;
        top--;
    } 
}
int T;
int main(){
    while(scanf("%d%d",&n,&m)!=EOF){
        top=0;tot=0;sumcol=0;tim=0;
        ans=0;bns=0;
        memset(to,0,sizeof(to));
        memset(low,0,sizeof(low));
        memset(dfn,0,sizeof(dfn));
        memset(vis,0,sizeof(vis));
        memset(col,0,sizeof(col));
        memset(net,0,sizeof(net));
        memset(out,0,sizeof(out));
        memset(into,0,sizeof(into));
        memset(from,0,sizeof(from));
        memset(stack,0,sizeof(stack));
        memset(visstack,0,sizeof(visstack));
        for(int i=1;i<=m;i++){
            int a,b;
            cin>>a>>b;
            add(a,b);
        }
        for(int i=1;i<=n;i++)
            if(!vis[i])    tarjin(i);
        if(sumcol==1){
            cout<<"0"<<endl;
            continue;
        }
        for(int i=1;i<=tot;i++)
            if(col[to[i]]!=col[x[i]]){
                into[col[to[i]]]++;
                out[col[x[i]]]++;
            }
        for(int i=1;i<=sumcol;i++){
            if(into[i]==0)    ans++;
            if(out[i]==0)    bns++;
        }
        cout<<max(ans,bns)<<endl;
    }
}