cf 1000c Covered Points Count_you are given n n segments on a coordinate line; e-优快云博客

本文链接：https://blog.youkuaiyun.com/max_kibble/article/details/80848918

本文介绍了一道CF1000C的线段覆盖问题，通过将线段的端点排序并扫描，计算每个整数坐标点被多少条线段覆盖，进而统计被k条线段覆盖的点的数量。

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一原题

You are given $n$ segments on a coordinate line; each endpoint of every segment has integer coordinates. Some segments can degenerate to points. Segments can intersect with each other, be nested in each other or even coincide.

Your task is the following: for every $k \in [1.. n]$ , calculate the number of points with integer coordinates such that the number of segments that cover these points equals $k$ . A segment with endpoints $l_{i}$ and $r_{i}$ covers point $x$ if and only if $l_{i} \leq x \leq r_{i}$ .

Input

The first line of the input contains one integer $n$ ( $1 \leq n \leq 2 \cdot 10^{5}$ ) — the number of segments.

The next $n$ lines contain segments. The $i$ -th line contains a pair of integers $l_{i}, r_{i}$ ( $0 \leq l_{i} \leq r_{i} \leq 10^{18}$ ) — the endpoints of the $i$ -th segment.

Output

Print $n$ space separated integers $c n t_{1}, c n t_{2}, \dots, c n t_{n}$ , where $c n t_{i}$ is equal to the number of points such that the number of segments that cover these points equals to $i$ .

  Examples
 

    input
    
     Copy
    
3
0 3
1 3
3 8

    output
    
     Copy
    
6 2 1 

    input
    
     Copy
    
3
1 3
2 4
5 7

    output
    
     Copy
    
5 2 0 

Note

The picture describing the first example:

$\text{[math]}$

Points with coordinates $[0, 4, 5, 6, 7, 8]$ are covered by one segment, points $[1, 2]$ are covered by two segments and point $[3]$ is covered by three segments.

The picture describing the second example:

$\text{[math]}$

Points $[1, 4, 5, 6, 7]$ are covered by one segment, points $[2, 3]$ are covered by two segments and there are no points covered by three segments.

二分析

给你n条x轴上的线段，问被k条线段覆盖的点的个数，k从1遍历到n。n<=2e5，线段坐标long long范围。

把2n个端点排序，相同的情况下起点在前，维护一下当前区间上有多少条线段（一个整型变量cnt，遇到起点+1，遇到终点-1），比较麻烦的是端点本身有多少条线段覆盖：如果遇到连续的一组重合的点，因为排序时起点在前，遍历完其中所有起点时cnt的值就是覆盖这个端点的线段数。

关于这种排序后扫一遍的题，cf上有篇看起来不错的博客，有空学习一下

PS：这题第一次提交的时候RE了，一开始以为是数组越界，而ans数组显然不会超。后来发现居然是sort的比较函数写的有毛病。关键就在于sort重载的是小于比较，两个相等的数比较应当返回false，附上错误的cmp：

bool cmp(const plb &p1, const plb &p2) {
	if (p1.fi == p2.fi) return p1.se; // first相同，second为true的排在前面
	return p1.fi < p2.fi; // first小的排在前面
}

三代码

/*
AUTHOR: maxkibble
LANG: c++
PROB: cf 1000c
*/

#include <bits/stdc++.h>

using namespace std;

#define fi first
#define se second
#define plb pair<long long, bool>

const int maxn = 2e5 + 5;

int n, cnt;
long long l, r, ans[maxn];
vector<plb> v;

bool cmp(const plb &p1, const plb &p2) {
	if (p1.fi == p2.fi) return (p1.se == p2.se)? false: p1.se;
	return p1.fi < p2.fi;
}

int main() {
	ios::sync_with_stdio(false);
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> l >> r;
		v.push_back(make_pair(l, true));
		v.push_back(make_pair(r, false));
	}
	sort(v.begin(), v.end(), cmp);
	bool ed = false;
	cnt = 1;
	for (int i = 1; i < v.size(); i++) {
		if (v[i].fi != v[i - 1].fi) {
			ans[cnt] += v[i].fi - (v[i - 1].fi) - 1;
			if (!ed) ans[cnt]++;
			else ed = false;
		}
		if (!ed && !v[i].se) {
			ans[cnt]++;
			ed = true;
		}
		cnt += v[i].se? 1: -1;
	}
	for (int i = 1; i <= n; i++) cout << ans[i] << " \n"[i == n];
	return 0;
}