树状数组-Stars

题目描述：

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.

 

解题感想：刚开始总是以为这是一道二维树状数组题，但其实是一道再简单不过的一维树状数组了，注意的是每次输入一颗星星的位置的时候，都要更新一次数据，这样可以使每次更新数据的时候都是站在y最大的角度，而不用考虑y的问题，每次输入一次就将对应的x的位置上的星星数加一就够了。下面是具体的AC代码：

 

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int star[32005],leave[15005];
int add(int x) {
 x+=1;
 for(int i=x;i<32005;i+=(i&-i))
 star[i]++;
}
int sum(int x) {
 x+=1;
 int sum=0;
 for(int i=x;i>0;i-=(i&-i))
 sum+=star[i];
 return sum;
}
int main() {
 int N;
 while(scanf("%d",&N)!=EOF) {
  int x,y;
  memset(star,0,sizeof(star));
  memset(leave,0,sizeof(leave));
  for(int i=1;i<=N;++i) {
   scanf("%d%d",&x,&y);
   add(x);
   leave[sum(x)-1]++;
  }
  for(int i=0;i<N;++i)
  printf("%d\n",leave[i]);
 }
 return 0;
}