Pku2352 Stars 树状数组

最新推荐文章于 2024-07-31 13:38:32 发布

转载最新推荐文章于 2024-07-31 13:38:32 发布 · 498 阅读

文章标签：

#iostream #input #output #numbers #each #file

本文介绍了一个天文学领域的算法问题，即如何统计二维平面上星星的分布层级。文章提供了一种利用离线数据结构来高效计算每个星星所在层级的方法，并附带了具体的C++实现代码。

Stars

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 22795		Accepted: 9933

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

代码一定要先查询，在更新。查询的即是小于等于x的点有多少个，就是该点的层数值。

代码：

#include <iostream>   
#include <cstdio>   
#include <string.h>   
using namespace std;  
int lev[32005];  
int sum1[32005];  
int lowbit(int x){  
  return x&(-x);  
}  
void update(int x){  
    while(x<32002){  
      sum1[x]++;  
      x+=lowbit(x);  
    }  
}  
int sum(int x){  
  int s=0;  
  while(x>0){  
    s+=sum1[x];  
    x-=lowbit(x);  
  }  
  return s;  
}  
  
int main(){  
  int n;  
  while(~scanf("%d",&n)){  
    int x,y,m;  
    m=n;  
    //printf("%d\n",n);   
    memset(lev,0,sizeof(lev));  
    memset(sum1,0,sizeof(sum1));  
    while(n--){  
      scanf("%d%d",&x,&y);  
      ++lev[sum(x+1)];  
      update(x+1);  
    }  
    for(int i=0;i<=m-1;++i)  
        printf("%d\n",lev[i]);  
  }  
  return 0;  
}