树状数组 POJ 2352 HDU 1541 Stars

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

51 15 17 13 35 5

Sample Output

12110

因为这个题目所给的输入很特殊，所以导致题目做起来很简单。

就是把一维的c数组不停的刷，相当于一行星星对应于c从左往右刷一遍。

这个和0-1背包问题的空间优化非常像，点击打开我的博客

代码：

#include<iostream>using namespace std;int n = 32001;int c[320005];int level[15001];int sum(int i){int s = 0;while (i){s += c[i];i -= (i&(-i));}return s;}void add(int i, int x){while (i <= n){c[i] += x;i += (i&(-i));}}int main(){int nn, s = 0, x, y;scanf("%d", &nn);for (int i = 0; i < nn; i++){scanf("%d%d", &x, &y);add(x + 1, 1);level[sum(x + 1) - 1]++;}for (int i = 0; i < nn; i++)printf("%d\n", level[i]);return 0;}

很明显可以看到，y在输入之后就没了，不做任何处理，也不调用y的值，很有意思吧

我还试过把n进行优化，然而这是不行的，n必须一开始就是最大值。

如果n从0开始，用x往上推的话，算出来的答案是错的。

而且，只要空间够大，本题lon n的计算时间是很小的，只有输入输出是θ（n）的时间。