CodeForces - 762E：Radio stations （CDQ分治||排序二分）

In the lattice points of the coordinate line there are n radio stations, the i-th of which is described by three integers:

x_i — the coordinate of the i-th station on the line,
r_i — the broadcasting range of the i-th station,
f_i — the broadcasting frequency of the i-th station.

We will say that two radio stations with numbers i and j reach each other, if the broadcasting range of each of them is more or equal to the distance between them. In other words min(r_i, r_j) ≥ |x_i - x_j|.

Let's call a pair of radio stations (i, j) bad if i < j, stations i and j reach each other and they are close in frequency, that is, |f_i - f_j| ≤ k.

Find the number of bad pairs of radio stations.

Input

The first line contains two integers n and k (1 ≤ n ≤ 10⁵, 0 ≤ k ≤ 10) — the number of radio stations and the maximum difference in the frequencies for the pair of stations that reach each other to be considered bad.

In the next n lines follow the descriptions of radio stations. Each line contains three integers x_i, r_i and f_i (1 ≤ x_i, r_i ≤ 10⁹, 1 ≤ f_i ≤ 10⁴) — the coordinate of the i-th radio station, it's broadcasting range and it's broadcasting frequency. No two radio stations will share a coordinate.

Output

Output the number of bad pairs of radio stations.

Examples

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Output

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https://www.cnblogs.com/mlystdcall/p/6351344.html

【转】：

首先整理一下条件：

每个点有三个属性，x,r,f

统计有多少对点i,j满足 min(ri,rj) >= |xi-xj| 且 |fi-fj| <= k，这样的点对被称作是“坏的”

对r值取min是个烦人的条件，于是我们把点按照r值从大到小排序，按照r值从大到小的顺序依次考虑每个点

这样对于每个点，我们只考虑它之前出现的点，也就是r值比他大的点，和他能不能组成“坏的”点对

这样的话，因为一个点i之前所有的点j的r值都比他大，所以 min(ri,rj) = ri

然后我们重新看一下问题：

按照指定的顺序依次加入点，每次加入一个点i，考虑它之前加入的所有点j，有多少个点满足 |xi-xj| <= ri 且 |fi-fj| <= k

再转化一下：

对于每个点i，考虑有多少个它之前的点j满足 xi-ri <= xj <= xi+ri 且 fi-k <= fj <= fi+k

我们把x和f这两个属性看做二维平面中的横纵坐标，问题就变成了：

向一个平面中添加一个点，查询指定矩形内点的个数

这是一个经典的三维偏序问题，可以用线段树套线段树或者 CDQ分治来做

---------------------------------------------分界线--------

自己：

注意到K比较小，我们也可以用二分暴力求矩形内部有多少个点。

（本来像练习CDQ分治的，但是太懒了，所以暴力了。

而且学到了维护有序的vector的方法，也不错的。

#include<bits/stdc++.h>
using namespace std;
const int maxn=100020;
vector<int>G[10020];
struct in{ int x,r,f; }s[maxn];
bool cmp(in w,in v){ return w.r>v.r; }
long long ans;
int main()
{
    int N,K,i,j;
    scanf("%d%d",&N,&K);
    for(i=1;i<=N;i++) scanf("%d%d%d",&s[i].x,&s[i].r,&s[i].f);
    sort(s+1,s+N+1,cmp);
    for(i=1;i<=N;i++){
        for(j=max(1,s[i].f-K);j<=s[i].f+K;j++)
             ans+=upper_bound(G[j].begin(),G[j].end(),s[i].x+s[i].r)-lower_bound(G[j].begin(),G[j].end(),s[i].x-s[i].r);
        int pos=upper_bound(G[s[i].f].begin(),G[s[i].f].end(),s[i].x)-G[s[i].f].begin();
        G[s[i].f].insert(G[s[i].f].begin()+pos,s[i].x);
    }
    printf("%I64d\n",ans);
    return 0;
}

CodeForces - 762E：Radio stations （CDQ分治||排序二分）

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