POJ2352 Stars 树状数组 + 裸

1）

为了避免出现0，保证从1开始，所以每次输入x以后，都先x++;

观察得知输入的数据因为有y 、x一次递增的规律，所以利用这一特性，只比较x。

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;const int maxn=32010;int star[maxn];int level[maxn];int Init(int x){        return x&(-x);//2^k,k是x以二进制形式时末尾0的个数}int Sum(int x){       int res=0;        while(x>0){                res+=star[x];                x-=Init(x);       }       return res;}void Add(int x){        while(x<=maxn){                star[x]++;                x+=Init(x);        }        return ;}int main(){        int n;        while(~scanf("%d",&n)){                memset(star,0,sizeof(star));                memset(level,0,sizeof(level));                int x;                int y;                for(int i=1;i<=n;i++){                        scanf("%d%d",&x,&y);                        x++;//保证从1开始，避免出现0                        level[Sum(x)]++;                        Add(x);//先计数，后加入，保证自身不是自己的子星星，而在自己正下方的点又能被包括进去。                }                for(int i=0;i<=n-1;i++){                        cout<<level[i]<<endl;                }        }        return 0;}

2）

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

51 15 17 13 35 5

Sample Output

12110

Hint

This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.