Educational Codeforces 58 (Rated for Div. 2) C. Division and Union (排序,二分)

题目连接: https://codeforces.com/problemset/problem/1101/C

题目大意: 给n条线段,求一个分割点,将线段分为两个集合,不同集合的线段不会有交点.

自己的做法是 将端点离散化,枚举端点来判断是否为分割点
怎么快速判断是否为分割点呢? 用一个数组sr记录右端点,一个数组sl记录左端点,对当前长度i,二分找到i在sr和sl中的位置,相当于找到多少个线段的右端点r<=i, 多少线段的左端点 l>i 然后判断和是否为n就可以了,注意集合要求非空 这样做复杂度好像比较高

#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<iostream>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<stdlib.h>
#include<algorithm>
#include<time.h>
#include<unordered_map>
#define bug1(g) cout<<"test: "<<g<<endl
#define bug2(g,i) cout<<"test: "<<g<<" "<<i<<endl
#define bug3(g,i,k) cout<<"test: "<<g<<" "<<i<<" "<<k<<endl
#define bug4(a,g,i,k) cout<<"test: "<<a<<" "<<g<<" "<<i<<" "<<k<<endl
using namespace std;
typedef long long ll;
const int maxn=2e5+5;
int t;
int n;
int l[maxn],r[maxn];
set<int> a;
vector<int>sl,sr;
int main()
{
    ios::sync_with_stdio(0);
    cin>>t;
    while(t--)
    {
        cin>>n;
        a.clear();
        sl.clear();
        sr.clear();
        for(int i =0;i<n;i++)
            {
                cin>>l[i]>>r[i];
                a.insert(l[i]);
                a.insert(r[i]);
                sl.push_back(l[i]);
                sr.push_back(r[i]);
            }
            sort(sl.begin(),sl.end());
            sort(sr.begin(),sr.end());
        int ans=-1;
        for(auto i=a.begin();i!=a.end();i++)
        {
            int r=upper_bound(sr.begin(),sr.end(),*i)-sr.begin(); // 一定都要用upper_bound 可以自己举例推一下
            int l=sl.end()-upper_bound(sl.begin(),sl.end(),*i);
            if(r>0&&l>0&&r+l==n)
            {
                ans=*i;
                break;
            }
        }
        if(ans==-1) cout<<-1<<endl;
        else
        {
            for(int i =0;i<n;i++)
            {
                if(r[i]<=ans) cout<<1;
                else cout<<2;
                if(i==n-1) cout<<endl;
                else cout<<" ";
            }
        }

    }
    return 0;
}

看巨巨博客还有一种更简单的做法 https://blog.youkuaiyun.com/CaprYang/article/details/86419347 排序后,直接记录之前出现过的最大的r与当前的l比较就行 自己又想复杂了()

#include <stdio.h>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int INF = 0x3f3f3f3f;
const int MAXN = 2e5 + 10;

struct node
{
	int l, r, i;
	bool operator < (const node &oth) const
	{
		return l < oth.l;
	}
}a[MAXN];
bool cmp(const node &a, const node &b)
{
	return a.i < b.i;
}
int main()
{
#ifdef LOCAL
	freopen("C:/input.txt", "r", stdin);
#endif
	int T;
	cin >> T;
	while (T--)
	{
		int n;
		cin >> n;
		for (int i = 1; i <= n; i++)
			scanf("%d%d", &a[i].l, &a[i].r), a[i].i = i;
		sort(a + 1, a + n + 1); //按照l排序
		int k = 0, r = a[1].r; //最远的r
		for (int i = 2; i <= n; i++)
		{
			if (a[i].l > r) //当前位置没覆盖到之前的r
				k = a[i].l; //记录可行的l
			r = max(r, a[i].r);
		}
		sort(a + 1, a + n + 1, cmp);
		if (k)
			for (int i = 1; i <= n; i++) //大于等于k所记录的分为第二组
				cout << (a[i].l >= k ? 2 : 1) << " ";
		else
			cout << -1;
		cout << endl;
	}

	return 0;
}