[ABC221D] Online games 题解

合集 - 题解(22)

1.AT_abc333_e [ABC333E] Takahashi Quest 题解2023-12-172.[ABC265F] Manhattan Cafe 题解2023-12-243.[ABC271E] Subsequence Path 题解2024-01-034.[ABC273D] LRUD Instructions 题解2024-01-055.P8085 [COCI2011-2012#4] KRIPTOGRAM 题解2024-02-296.[ABC238F] Two Exams 题解2024-02-297.[ABC217F] Make Pair 题解2024-03-018.[ABC219F] Cleaning Robot 题解2024-03-099.[ABC219E] Moat 题解2024-03-10

10.[ABC221D] Online games 题解2024-04-04

11.[ABC221E] LEQ 题解2024-04-0412.[ABC223E] Placing Rectangles 题解2024-04-0413.[ABC211D] Number of Shortest paths 题解2024-04-0414.[ABC211F] Rectilinear Polygons 题解2024-04-0415.[ABC223F] Parenthesis Checking 题解2024-04-0416.CF154C Double Profiles 题解2024-04-1617.[ABC208D] Shortest Path Queries 2 题解2024-04-1718.[ABC212E] Safety Journey 题解2024-04-1719.[ABC229E] Graph Destruction 题解2024-04-1720.[ABC240E] Ranges on Tree 题解2024-04-1821.[ABC261E] Many Operations 题解2024-05-1222.P10842 【MX-J2-T3】Piggy and Trees 题解2024-08-05

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[ABC221D] Online games 题解

思路解析

可以发现题目就是单纯区间加和查询每一个值有多少次出现。首先看到区间修改加结束时查询可以想到差分，但是通过 $A_i\le {10}^9$ 发现值域很大没法直接根据值差分。于是想到离散化，将每个点离散下来，统计每两个离散的时间之间在线的人乘上时间段的长度即可。

code

#include<bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
int n, a[N], b[N], c[5 * N], ans[N];
int main() {
	cin >> n;
	vector<int> v;
	for(int i = 1; i <= n; i++) {
		cin >> a[i] >> b[i];
		v.push_back(a[i]);
		v.push_back(a[i] + b[i]);
	}
	v.push_back(n + 1);
	sort(v.begin(), v.end());
	v.erase(unique(v.begin(), v.end()), v.end());	//离散化去重
	for(int i = 1; i <= n; i++) {
		int pa = lower_bound(v.begin(), v.end(), a[i]) - v.begin() + 1;
		int pb = lower_bound(v.begin(), v.end(), a[i] + b[i]) - v.begin() + 1;
		c[pa]++; c[pb]--;	//差分
	}
	for(int i = 1; i < v.size(); i++) {
		c[i] += c[i - 1];
		ans[c[i]] += v[i] - v[i - 1];	//统计区间
	}
	for(int i = 1; i <= n; i++) {
		cout << ans[i] << ' ';
	}
	return 0;
}