POJ2352 Stars

最新推荐文章于 2021-05-03 20:08:45 发布

转载最新推荐文章于 2021-05-03 20:08:45 发布 · 64 阅读

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原文链接：http://www.cnblogs.com/SilverNebula/p/5836963.html

本文介绍了一个天文学问题的算法解决方案，即统计星图中各星等的分布情况。利用树状数组来高效处理大量坐标数据，确保算法在限定时间内完成。输入数据包括星星数量及它们的坐标，输出为各星等的星星数量。

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 42670		Accepted: 18577

Description

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.

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 file 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

Sample Output

Hint

This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.

Source

Ural Collegiate Programming Contest 1999

树状数组裸题……

输入保证y递增，连排序都省了。

 1 /**/
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<cmath>
 5 #include<cstring>
 6 #include<algorithm>
 7 using namespace std;
 8 const int mxn=50000;
 9 long long t[mxn];
10 int n,m,k;
11 int a[mxn];
12 int r[mxn];
13 inline int lowbit(int x){
14     return x&-x;
15 }
16 void add(int p,int v){
17     while(p<=mxn){
18         t[p]+=v;
19         p+=lowbit(p);
20     }
21     return;
22 }
23 int sum(int p){
24     int res=0;
25     while(p){
26         res+=t[p];
27         p-=lowbit(p);
28     }
29     return res;
30 }
31 int main(){
32     scanf("%d",&n);
33     int i,j;
34     int x,y;
35     for(i=1;i<=n;i++){
36         scanf("%d%d",&x,&y);
37         x++;
38         r[sum(x)]++;
39         add(x,1);
40     }
41     for(i=0;i<n;i++)printf("%d\n",r[i]);
42     return 0;
43 }

转载于:https://www.cnblogs.com/SilverNebula/p/5836963.html