hdu3836Equivalent Sets-优快云博客

本文链接：https://blog.youkuaiyun.com/imzxww/article/details/79198919

Equivalent Sets

Time Limit: 12000/4000 MS (Java/Others) Memory Limit: 104857/104857 K (Java/Others)
Total Submission(s): 5277 Accepted Submission(s): 1916

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

Sample Output

4
2
思路：首先求出图的所有强连通分量,然后缩点成DAG图.求出新图中所有点的出度=0与入度=0的节点数大的即为所求需要添加的边.
代码：
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <cmath>
using namespace std;
const int N=20000+10;
int dfn[N];
int low[N];
int belong[N];
int in[N];
int out[N];
bool instack[N];
stack<int>s;
int n,m;
int cnt,index;
vector<int>e[N];
void init()
{
    int i;
    for(i=0;i<=n;i++)
        e[i].clear();
    while(!s.empty())
        s.pop();
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(belong,0,sizeof(belong));
    memset(instack,false,sizeof(instack));
    index=cnt=0;
    memset(in,0,sizeof(in));
    memset(out,0,sizeof(out));
}
void tarjan(int u)
{
    dfn[u]=low[u]=++index;
    int i;
    instack[u]=true;
    s.push(u);
    int v;
    for(i=0;i<e[u].size();i++)
    {
        v=e[u][i];
        if(!dfn[v])
        {
            tarjan(v);
            low[u]=min(low[v],low[u]);
        }
        else if(instack[v])
            low[u]=min(dfn[v],low[u]);
    }
    if(dfn[u]==low[u])
    {
        cnt++;
        do
        {
            v=s.top();
            s.pop();
            belong[v]=cnt;
            instack[v]=false;
        }
        while(u!=v);
    }
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        int i,j;
        init();
        int a,b;
        for(i=0;i<m;i++)
        {
            scanf("%d%d",&a,&b);
            e[a].push_back(b);
        }
        for(i=1;i<=n;i++)
        {
            if(!dfn[i])
                tarjan(i);
        }
        for(i=1;i<=n;i++)
        {
            for(j=0;j<e[i].size();j++)
            {
                if(belong[i]!=belong[e[i][j]])
                {
                    out[belong[i]]++;
                    in[belong[e[i][j]]]++;
                }
            }
        }
        int ans=0,ans1=0;
        for(i=1;i<=cnt;i++)
        {
            if(in[i]==0)ans++;
            if(out[i]==0)ans1++;
        }
        if(cnt==1)printf("0\n");
        else printf("%d\n",max(ans,ans1));
    }
    return 0;
}