HDU 1541 Stars（树状数组）

Stars

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 9546 Accepted Submission(s): 3823

Problem 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 ou
4000
tput 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

题意：计算每颗星星左下边包括了多少颗星星，这个数值就是等级。不超过本身的x，y的坐标。问每种等级有多少颗星星。

题解：题目给的数据按照y排序，相同的y按照x排序，先看x坐标的星星，只看y值相同的星星坐标的x。当一个x输入时，它之后的所有星星的等级肯定都要加一。这点就是树状数组的最基本应用-单点更新。然后get_sum区间求和，它的等级就等于前面已经输入的x坐标在[0,x]区间的星星数量。

代码：

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#pragma comment(link,"/STACK: 102400000,102400000")
using namespace std;
const int maxn=32005;
int c[maxn],_rank[maxn];
int lowbit(int x)///求的是某个点管辖范围；
{
return x&(-x);
}
int get_sum(int  x)///1~x的区间和
{
int sum=0;
while(x)
{
sum+=c[x];
x-=lowbit(x);
}
return sum;
}
void add(int i,int x)///用树状数组对后面节点更新
{
while(i<=x)
{
c[i]++;
i +=lowbit(i);
}
}

int main()
{
int n;
while(cin>>n)
{
memset(c,0,sizeof(c));
memset(_rank,0,sizeof(_rank));
int x,y;
for(int i=0;i<n;i++)
{
cin>>x>>y;
_rank[get_sum(x+1)]++;
add(x+1,maxn);
}
for(int i=0;i<n;i++)
cout<<_rank[i]<<endl;
}
return 0;
}