POJ2352 Stars 树状数组

最新推荐文章于 2022-03-19 16:31:18 发布

verdin黄大锤

最新推荐文章于 2022-03-19 16:31:18 发布

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本文链接：https://blog.youkuaiyun.com/rabbit_ZAR/article/details/58294167

本文介绍了一个天文学领域的算法问题，即如何通过星星的坐标来计算每个星星的等级，并统计各个等级星星的数量。文章提供了一种使用树状数组进行高效计算的方法。

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题目：

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

题目大意：

一张图上有一些星星，这些星星左上角的星星（横纵坐标都小于该星星的坐标）数量就是它们的等级。

给定星星们的坐标，输出n行，每一行i（0<=i<n）代表等级为i的星星数量。

思路：

因为为题目中说，按照x为第一关键字，y为第二关键字按顺序输入，所以在计算时不必考虑y。所以将当前已经输入的点投影到x轴上，当前点x的等级就是它上方点的数量。用树状数组求解。

代码：

#include<cstdio>
#include<iostream>
using namespace std;

int n;
int a[32010]= {0};

int lowbit(int x) {
	return x&(-x);
}

void add(int x) {
	while(x<=32005) {
		a[x]++;
		x+=lowbit(x);
	}
}

int getsum(int x) {
	int s=0;
	while(x>0) {
		s+=a[x];
		x-=lowbit(x);
	}
	return s;
}

int main() {
	scanf("%d",&n);

	int sum[15010]= {0};

	for(int i=1; i<=n; i++) {
		int x,y;
		scanf("%d%d",&x,&y);
		x++,y++;
		sum[getsum(x)]++;
		add(x);
	}
	
	for(int i=0;i<n;i++){
		printf("%d\n",sum[i]);
	}
	
	return 0;
}