【树状数组】Pku 2325 Stars_2325stars-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_43398760/article/details/93333627

博客围绕星星坐标统计问题展开，天文学家想了解星星等级分布，即每颗星星左下角星星数量。输入星星坐标，按特定顺序排列，要求编写程序统计各等级星星数量。此问题可利用树状数组进行基础的统计计数。

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Description

Astronomers often examine star maps where stars are represented by points on a plane and each star h
as 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 s
tars.

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 th
e 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.
输入数据按y值排序好了，如果y值一样则按x值升序。注意x的值可能为0

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

分析

题目大意

给你一堆星星的坐标，对于每颗星星求出在它左下角的星星数量

这道题比较基础，利用树状数组进行统计计数就好

代码

#include<bits/stdc++.h>
using namespace std;
int n;
int tree[32010];
int level[32010];
 
inline int read() {
    int x = 0, f = 1; char ch = getchar();
    while (ch < '0' || ch > '9') {if (ch == '-') f = -1; ch = getchar();}
    while (ch >= '0' && ch <= '9') {x = (x << 1) + (x << 3) + (ch ^ 48); ch = getchar();}
    return x * f;
}
 
int ask(int x) {
    int sum = 0;
    for (; x; x -= (x & -x)) {
        sum += tree[x];
    }
    return sum;
}
 
void add(int x, int y) {
    for (; x <= 32000; x += (x & -x)) {
        tree[x] += y;
    }
}
 
int main() {
    n = read();
    for (int i = 1; i <= n; i++) {
        int x = read() + 1, y = read();
        level[ask(x)]++;
        add(x, 1);
    }
    for (int i = 0; i < n; i++) printf("%d\n", level[i]);
}