POJ 2352 (树状数组)

C - Stars

Crawling in process...Crawling failedTime Limit:1000MS    Memory Limit:65536KB    64bit IO Format:%lld & %llu

SubmitStatus

Description

Input

Output

Sample Input

Sample Output

Hint

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

Hint

This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.

题目大意：

在坐标上有n个星星，如果某个星星坐标为(x, y), 它的左下位置为：
(x0,y0)，x0&lt;=x 且y0&lt;=y。如果左下位置有a个星星，就表示这个星
星属于level x

按照y递增，如果y相同则x递增的顺序给出n个星星，求出所有level水平的数量。

分析与总结：
因为输入是按照按照y递增，如果y相同则x递增的顺序给出的， 所以，对于第i颗星星，它
的level就是之前出现过的星星中，横坐标x小于等于i星横坐标的那些星星的总数量（前面
的y一定比后面的y小）。
所以，需要找到一种数据结构(树状数组)来记录所有星星的x值，方便的求出所有值为0～x的星星总数量。

注意：有大量IO操作，使用scanf()和printf()。

#include<iostream>using namespace std;#define lowbit(x) (x&-x)#define N 32005int n;int c[N],ans[N];void add(int x,int sum) {while (x <= N) {c[x] += sum;x += lowbit(x);}}int sum(int x) {//x!=0;int s=0;while (x > 0) {s += c[x];x -= lowbit(x);}return s;}int main() {cin >> n;int x, y;for (int i = 0; i < n; i++) {cin >> x >> y;x++;ans[sum(x)]++;add(x, 1);}for (int i = 0; i < n; i++)cout << ans[i] << endl;return 0;}