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POJ 2352 Stars(树状数组)

时间:2015-04-22 22:22:01      阅读:180      评论:0      收藏:0      [点我收藏+]

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Stars
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 35467   Accepted: 15390

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. 
技术分享

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

Hint

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

Source





         题意:问在输入的点(星星)的坐标的左下方(包含左方和下方)有几个点(星星)。有几个点(星星)就是第几种情况,一共有1~n种情况,输出每种情况的个数。因为题目保证y的坐标是按升序输入的,所以就不用考虑y,只需要更新x的树状数组就可以了。


#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

using namespace std;

const int maxn = 32001;

int c[maxn];
int v[maxn];
int n;

int lowbit(int x)
{
    return x&(-x);
}

void updata(int i,int k)
{
    while(i<=maxn)
    {
        c[i] += k;
        i += lowbit(i);
    }
}

int getsum(int i)
{
    int sum = 0;
    while(i>0)
    {
        sum += c[i];
        i -= lowbit(i);
    }
    return sum;
}

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        int x,y;
        memset(v,0,sizeof(v));
        memset(c,0,sizeof(c));
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&x,&y);
            x++;
            v[getsum(x)]++;
            updata(x,1);
        }
        for(int i=0;i<n;i++)
        {
            printf("%d\n",v[i]);
        }
    }
    return 0;
}



POJ 2352 Stars(树状数组)

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原文地址:http://blog.csdn.net/yeguxin/article/details/45199347

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