杭电oj 1541 树状数组

暑假时和xd讨论过这道题，当时迷迷糊糊的做了出来，昨天做专题练习时，怎么也想不出来怎么和树状数组联系起来。昨天想了有几个小时，今天又想了想，终于想了出来。自己感觉这是一道树状数组的好题，也是有一定难度的。

题目：

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

ac代码：

代码一定要先查询，在更新。查询的即是小于等于x的点有多少个，就是该点的层数值。

#include <iostream>
#include <cstdio>
#include <string.h>
using namespace std;
int lev[32005];
int sum1[32005];
int lowbit(int x){
  return x&(-x);
}
void update(int x){
	while(x<32002){
	  sum1[x]++;
	  x+=lowbit(x);
	}
}
int sum(int x){
  int s=0;
  while(x>0){
    s+=sum1[x];
	x-=lowbit(x);
  }
  return s;
}

int main(){
  int n;
  while(~scanf("%d",&n)){
    int x,y,m;
	m=n;
	//printf("%d\n",n);
	memset(lev,0,sizeof(lev));
	memset(sum1,0,sizeof(sum1));
	while(n--){
	  scanf("%d%d",&x,&y);
	  ++lev[sum(x+1)];
	  update(x+1);
	}
	for(int i=0;i<=m-1;++i)
		printf("%d\n",lev[i]);
  }
  return 0;
}