
【Ural_P1028】Stars
发布日期:2021-05-06 16:00:32
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分类:技术文章
本文共 1812 字,大约阅读时间需要 6 分钟。
Stars
Astronomers often examine star maps where stars are represented by points on a plane and each star has Cartesian coordinates. Let the level of a star be an amount of the stars that are not higher and not to the right of the given star. Astronomers want to know the distribution of the levels of the stars.
Problem illustration For example, look at the map shown on the figure above. Level of the star number 5 is equal to 3 (it’s formed by three stars with a numbers 1, 2 and 4). And the levels of the stars numbered by 2 and 4 are 1. At this map there are only one star of the level 0, two stars of the level 1, one star of the level 2, and one star of the level 3. You are to write a program that will count the amounts of the stars of each level on a given map.Input
The first line of the input contains a number of stars N (1 ≤ N ≤ 15000). The following N lines describe coordinates of stars (two integers X and Y per line separated by a space, 0 ≤ X,Y ≤ 32000). There can be only one star at one point of the plane. Stars are listed in ascending order of Y coordinate. Stars with equal Y coordinates are listed in ascending order of X coordinate.
Output The output should contain N lines, one number per line. The first line contains amount of stars of the level 0, the second does amount of stars of the level 1 and so on, the last line contains amount of stars of the level N−1.Sample
input
51 15 17 13 35 5
output
12110
解题思路
树状数组模板题
#include#include using namespace std;int n,c[15010],ans[15010];void in(int x){ for(;x<=32010;x+=x&(-x)) c[x]+=1;}int find(int x){ int s=0; for(;x;x-=x&(-x)) s+=c[x]; return s;}int main(){ cin>>n; for(int i=1;i<=n;i++) { int x,y; scanf("%d%d",&x,&y); x++; ans[find(x)]++; in(x); } for(int i=0;i<=n-1;i++) cout< <
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初次前来,多多关照!
[***.217.46.12]2025年03月31日 07时11分51秒
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