poj2352 Stars（树状数组）

Stars

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.

题意：定义星星的等级为在它左下角（包括正左和正下）的星星的个数。给出若干个星星的坐标（不超过15000个），输出各个等级的星星分别有多少个。每个星星的坐标不超过32000，输入时按照Y坐标升序输入。

分析：因为题目已经按照Y坐标升序排列了，只需要每次读入时用树状数组统计X坐标比小于等于X的星星有多少个算出其等级再统计就可以了。不过，要注意坐标有可能为0，为0时会死循环，所以在读入坐标时应加1。

代码

#include 
#include 
#include 
#define max 32001
using namespace std;

int c[max],ans[max],x,y,n;

int sum(int x)
{
    int s=0;
    while (x>0)
    {
        s+=c[x];
        x-=x&(-x);
    }
    return s;
}

void insert(int x)
{
    while (x<=max)
    {
        c[x]++;
        x+=x&(-x);
    }
}

int main()
{
    while (~scanf("%d",&n))
    {
    memset(c,0,sizeof(c));
    memset(ans,0,sizeof(ans));
    for (int i=1;i<=n;i++)
    {
        scanf("%d%d",&x,&y);
        x++;
        ans[sum(x)]++;
        insert(x);
    }
    for (int i=0;iprintf("%d\n",ans[i]);
    }
}