HDU-1541: Stars（树状数组）

树状数组

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

题意：输入n，接下来有n个坐标，输入的坐标按纵坐标非递减排列，如果纵坐标相等，则横坐标按递增排列，不会有相同的坐标。如果有n个点位于某一点的左下角（包括左边和正下方），则此点的等级为n，最后输出等级为0至n-1的点的数量。

分析：由于坐标是按y的递增给出的，那么就只需考虑水平方向，如果给出一个点的坐标为(x,y)，那么它的等级就等于前面已经输入的x坐标在[0,x]区间的点数量。

• 完整代码：

#include<stdio.h>
#include<string.h>
int n,a[32005],s[32005];
int lowbit(int x)
{
    return x&(-x);
}
void change(int x,int y) //在x点加上y
{
    for(int i=x;i<=32005;i+=lowbit(i))
        a[i]+=y;
}
int sum(int x) //1-x范围内求和
{
    int sum=0;
    for(int i=x;i>0;i-=lowbit(i))
        sum+=a[i];
    return sum;
}
int main()
{
    int i,x,y;
    while(scanf("%d",&n)!=EOF)
    {
        memset(a,0,sizeof(a));
        memset(s,0,sizeof(s));
        for(i=1;i<=n;i++)
        {
            scanf("%d%d",&x,&y);
            x++;  //x可能输入的是0,0&(-0)不正确
            s[sum(x)]++; //统计这个范围内的数量
            change(x,1); 
        }
        for(i=0;i<n;i++)
            printf("%d\n",s[i]);
    }
    return 0;
}