2020牛客暑期多校训练营(第八场)I.Interesting Computer Game(并查集)

题目链接

思路:

将题目转化为求连通子图的顶点数之和,若无环贡献为顶点数,否则为顶点数减1。(数字过大需要用到离散化优化)

代码:

#include<bits/stdc++.h>
#define int long long
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
const int N=2e5+10;
const int M=2e4+5;
const double eps=1e-8;
const int mod=1e9+7;
const int inf=0x7fffffff;
const double pi=3.1415926;
using namespace std;
pair<int,int>p[N];
int a[N<<1],es,ps,vis[N<<1],ans,s[N<<1],c[N<<1];
int find(int x)
{
    return x==s[x]?x:s[x]=find(s[x]);
}
signed main()
{
    IOS;
	int t;
	cin>>t;
	for(int k=1;k<=t;k++)
    {
		int n,tot=0;
		cin>>n;
		for(int i=1;i<=n;i++)
        {
			cin>>p[i].first>>p[i].second;
			a[++tot]=p[i].first,a[++tot]=p[i].second;
		}
		sort(a+1,a+tot+1);
		tot=unique(a+1,a+tot+1)-a-1;
		for(int i=1;i<=tot;i++)
		{
		    s[i]=i;
            c[i]=0;
		}
		for(int i=1;i<=n;i++)
        {
			int x=p[i].first,y=p[i].second;
			x=lower_bound(a+1,a+tot+1,x)-a;
			y=lower_bound(a+1,a+tot+1,y)-a;
			x=find(x),y=find(y);
			if(x!=y)
			{
			    s[x]=y,c[y]|=c[x];
			}
			else
			{
			    c[x]=1;
			}
		}
		ans=tot;
		for(int i=1;i<=tot;i++)
        {
            if(find(i)==i&&!c[i])
            {
                ans--;
            }
        }
		cout<<"Case #"<<k<<": "<<ans<<endl;
	}
	return 0;
}

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转载自blog.csdn.net/ACkingdom/article/details/107777517