hdu 1541 Stars 树状数组

Stars

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

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 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
3 3
7 1
5 5

 

Sample Output

1
2
1
1
0

题意是:先按y递增,y相同再按x递增的顺序 给你很多星星,每个星星的等级等于它左下角(包括x和y坐标相等)的星星总数
然后从等级0到n-1,输出每个等级对应的星星的个数

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

int maxn=32001;
int c[32001],ans[15001];
int lowbit(int x)
{
    return x&(-x);
}
void update(int x,int val)
{
    for(; x<=maxn; x+=lowbit(x)) c[x]+=val;
}

int sum(int x)
{
    int sum=0;
    for(; x>0; x-=lowbit(x)) sum+=c[x];
    return sum;
}

int main()
{
    int x,y,i,n;
    while(~scanf("%d",&n))
    {
        memset(c,0,sizeof(c));
        memset(ans,0,sizeof(ans));
        for(i=1; i<=n; i++)
        {
            scanf("%d %d",&x,&y);
            x++;
            ans[sum(x)]++;//由于是按y轴坐标递增的顺序给出的,所以此星星左下角的星星总数就是x轴坐标从1到x的每个坐标x 对应星星的数的总和,因为x可能等于0,所以每次输入的x都把它加一然后更新此点,于是sum(x)就是此星星等级
            update(x,1);
        }
        for(i=0; i<n; i++)
            printf("%d\n",ans[i]);
    }
    return 0;
}