CF1278D Segment Tree(set+并查集)

// Problem: CF1278D Segment Tree
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/CF1278D
// Memory Limit: 250 MB
// Time Limit: 2000 ms
// Author:Cutele
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll>PLL;
typedef pair<int, int>PII;
typedef pair<double, double>PDD;
#define I_int ll
inline ll read()
{
    ll x = 0, f = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
        if(ch == '-')f = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9')
    {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}
  
inline void out(ll x){
    if (x < 0) x = ~x + 1, putchar('-');
    if (x > 9) out(x / 10);
    putchar(x % 10 + '0');
}
  
inline void write(ll x){
    if (x < 0) x = ~x + 1, putchar('-');
    if (x > 9) write(x / 10);
    putchar(x % 10 + '0');
}
  
#define read read()
#define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define multiCase int T;cin>>T;for(int t=1;t<=T;t++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i<(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define perr(i,a,b) for(int i=(a);i>(b);i--)
ll ksm(ll a, ll b,ll p)
{
    ll res = 1;
    while(b)
    {
        if(b & 1)res = res * a%p;
        a = a * a %p;
        b >>= 1;
    }
    return res;
}

const int inf = 0x3f3f3f3f;
const int maxn=5e5+7;

int root[maxn];
struct node{
	int l,r;
	bool operator < (const node & A) const{
		return l < A.l;
	}
}a[maxn];



int Find(int x){
	if(x!=root[x]) root[x]=Find(root[x]);
	return root[x];
}
set<node>s;
int main(){
	int n=read;
	rep(i,1,n){
		a[i].l=read,a[i].r=read;root[i]=i;
	}
	sort(a+1,a+n+1);
	s.insert({a[1].r,1});
	rep(i,2,n){
		while(s.size()){
			node t=*s.begin();
			if(t.l<=a[i].l) s.erase(s.begin());
			else break;
		}
		for(set<node>::iterator it=s.begin();it!=s.end();it++){
			node t=*it;
			if(t.l<a[i].r){
				int u=i,v=t.r;
				int fu=Find(u),fv=Find(v);
				if(fu!=fv) root[fu]=fv;
				else{
					puts("NO");return 0;
				}
			}
			else break;
		}
		s.insert({a[i].r,i});
	}
	int res=0;
	for(int i=1;i<=n;i++)
		if(i==Find(i)) res++;
	if(res==1) puts("YES");
	else puts("NO");
	return 0;
}


原文地址:https://www.cnblogs.com/OvOq/p/15096411.html