[树状数组]POJ 2352 Stars

传送门：Stars

Stars

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 29248		Accepted: 12790

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

5
1 1
5 1
7 1
3 3
5 5

Sample Output

1
2
1
1
0

Hint

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

Source

Ural Collegiate Programming Contest 1999

解题报告：

算法有很多种,最实用的是树状数组
由于本题的数据输入是以y坐标的升序输入，所以，只需要比较x坐标即可
树状数组存储的是某一段范围内有多少个点，即求和.

代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#define N 15005
#define M 32005
using namespace std;
int ans[M],x[N],result[M];
int n,y,maxn;
int lowbit(int n){   //lowbit
    return n & (-n);
}
void modify(int x){  //修改modify
	for(int i = x; i<=maxn+1; i+=lowbit(i))
        ans[i]++;
}
int getSum(int x){  //求和sum
	int sum= 0;
	for (int i = x; i > 0; i -= lowbit(i))
		sum += ans[i];
	return sum;
}
int main(){
    int i;
    scanf("%d", &n);
    maxn = 0;
    for(i=1; i<=n; i++){
        scanf("%d%d", &x[i], &y);
        maxn = max(x[i],maxn);
    }
    memset(result, 0, sizeof(result));
    for(i=1; i<=n; i++){
        result[getSum(x[i]+1)]++;
        modify(x[i]+1);
    }
    for(i=0; i<n; i++)
        printf("%d\n", result[i]);
    return 0;
}