Stars--树状数组

Stars

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 44636 Accepted: 19380

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.

题目链接：http://poj.org/problem?id=2352

这个是我学树状数组做的第一题，本以为做了个例题，但是没想到还是训练计划里的一个题。

题目大意是说按y递增的顺序给出n颗星星的坐标(y相等则x递增)，每个星星的等级，等于在它左边且在它下边（包括水平和垂直方向）的星星的数量，求出等级为0到n-1的星星分别有多少个。

 

思路：树状数组。由于星星的坐标是按y的递增给出的，那么就只需考虑水平方向就好了，如果给出一个星星的坐标为(a,b)，那么它的等级就等于前面已经输入的x坐标在[0,a]区间的星星数量。

代码：

#include <cstdio>#include <cstring>#include <iostream>using namespace std;const int MAXN=32005;int c[MAXN],level[MAXN],n;int lowbit(int x){    return x&(-x);}int sum(int n){    int sum=0;    while(n>0){        sum+=c[n];        n-=lowbit(n);    }    return sum;}void add(int x){    while(x<=MAXN){        ++c[x];        x+=lowbit(x);    }}int main(){    int n,x,y;    while(~scanf("%d",&n)){        memset(level,0,sizeof(level));        memset(c,0,sizeof(c));        for(int i=0;i<n;i++){            scanf("%d%d",&x,&y);            ++x;            level[sum(x)]++;            add(x);        }        for(int i=0;i<n;i++){            printf("%d\n",level[i]);        }    }    return 0;}