树状数组——POJ2352

最新推荐文章于 2019-11-29 19:29:37 发布

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最新推荐文章于 2019-11-29 19:29:37 发布

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原文链接：http://www.cnblogs.com/Skyvot/p/4036495.html

本文介绍了一种用于计算星图中每个等级星星数量的算法。通过输入多个点的坐标，程序能够输出每个等级对应的星星数量。算法利用0、1数组记录坐标，并通过树状数组统计横坐标小于当前点的星星数量。

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水题，233

Stars

Time Limit: 1000MS
Memory Limit: 65536KTotal Submissions: 33189
Accepted: 14500

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.

题目大意：给你好几个点的坐标，问在该点左下方的区域内有k（1<=k<=n）点（不包括自己，包括横坐标或纵坐标相同的）的点有几个。

好了，这个题水到不能再水了。

我们只需要用0，1数组来记录坐标就好了，由于给你的点的坐标都是按照纵坐标排好序的，所以只需记录横坐标就好。0表示没有，1表示有，然后对于我们的每一个点，只要用树状数组统计一下比当前点横坐标x小的点有多少就好了。

不想写了，上代码（cin和cout会超时，自己看着办。。。。。）

 1 #include<iostream>
 2 using namespace std;
 3 int n,tree[33000];
 4 int level[33000];
 5 int lowbit(int x)
 6 {
 7     return x&(-x);
 8 }
 9 void add(int x,int k)
10 {
11     for(int i=x;i<=33000;i+=lowbit(i))
12     {
13         tree[i]+=k;
14     }
15 }
16 int get(int x)
17 {
18     int sum=0;
19     for(int i=x;i>0;i-=lowbit(i))
20     {
21         sum+=tree[i];
22     }
23     return sum;
24 }
25 int main()
26 {
27     cin>>n;
28     for(int i=1;i<=n;i++)
29     {
30         int x,y;
31         cin>>x>>y;
32         level[get(++x)]++;
33         add(x,1);
34     }
35     for(int i=0;i<n;i++)
36     {
37         cout<<level[i]<<endl;
38     }
39     return 0;
40 }