【HDU - 1541 - Stars 【区间求和单点更新 ==树状数组】

最新推荐文章于 2020-08-18 11:29:10 发布

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最新推荐文章于 2020-08-18 11:29:10 发布

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分类专栏：树状数组

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本文介绍了一个天文学领域的编程问题：如何统计星图中各星星的级别。通过使用一种特殊的数据结构，程序能够高效地计算每个星星周围不高于且不位于其右侧的星星数量。文章提供了一段C++代码示例，展示了如何实现这一功能。

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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.

这里写图片描述

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
5
1 1
5 1
7 1
3 3
5 5
Sample Output
1
2
1
1
0

读懂题目，就好写了，题目的输入也很友好。
代码

#include<stdio.h>
#include<algorithm> 
#include<math.h>
#include<string.h>
using namespace std;
#define  LL long long
#define fread() freopen("in.txt","r",stdin)
#define fwrite() freopen("out.txt","w",stdout)
#define CLOSE() ios_base::sync_with_stdio(false)

const int MAXN = 15000+10;
const int MAXM = 1e6+10;
const int mod = 1e9+7;
const int inf = 0x3f3f3f3f;
const int MAX = 32000+10;
int lowbit(int x) { return x&(-x);  }
int c[MAX];
int n;
int sum(int x){
    int s=0;
    while(x>0){   
        s+=c[x];
        x-=lowbit(x);
    }
    return s;
}

int add(int x,int val){
    while(x<=MAX){
        c[x]+=val;
        x+=lowbit(x);
    }
}
int a[MAXN];
int main(){
     while(scanf("%d",&n)!=EOF){
        memset(a,0,sizeof(a));
        memset(c,0,sizeof(c));
        for(int i=1;i<=n;i++)   {
            int x,y;scanf("%d%d",&x,&y);
            int z=sum(x+1);  //因为X可能为0，所以这里要+1  
            a[z]++;
            add(x+1,1); 
         } 
        for(int i=0;i<n;i++)  
            printf("%d\n",a[i]);
     }
    return 0;
}