zoukankan      html  css  js  c++  java
  • poj 2352 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.

    fd47938575a43734c7155490c64a749e

    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.


    树状数组水题,题目给的数据已经按照y优先的顺序排好序了,所以连排序都不用,直接边输入边计算当前点的level值。

    #include<iostream>
    #include<algorithm>
    #include<cstring>
    #include<cstdio>
    using namespace std;
    int C[32005],n;
    int lowbit(int x)
    {
        return x&-x;
    }
    int sum(int x)      //求前缀和,x向左上爬
    {
        int ret=0;
        while(x>0)
        {
            ret+=C[x];
            x-=lowbit(x);
        }
        return ret;
    }
    void add(int x,int d)     //A[x]加上d,x向右上爬
    {
        while(x<=n)
        {
            C[x]+=d;
            x+=lowbit(x);
        }
    }
    
    int ans[15005];
    int main()
    {
        int N,x,y;
        while(scanf("%d",&N)!=EOF)
        {
            n=32003;
            memset(C,0,sizeof(C));
            memset(ans,0,sizeof(ans));
            for(int i=1; i<=N; i++)
            {
                scanf("%d%d",&x,&y);
                x++;y++;
                ans[sum(x)]++;
                add(x,1);
            }
            for(int i=0;i<N;i++){
                printf("%d
    ",ans[i]);
            }
        }
    }
  • 相关阅读:
    在美国,一名 Uber 司机能赚多少?
    多名Uber司机被指刷单遭封号 一周薪水为0
    腾讯内推【腾讯工作机会内推】【腾讯社招内推】(长期有效)
    优步uber司机怎么注册不了?注册优步司机问题要点
    如何注册成为uber司机 快速成为优步司机网上注册流程攻略 2015最新
    Google AdSense的CPC点击单价超百度联盟(2014)
    非流动资产(财务报表解读)
    资产负债率
    流动资产(财务报表解读)
    UBER人民优步司机注册攻略
  • 原文地址:https://www.cnblogs.com/lastone/p/5297832.html
Copyright © 2011-2022 走看看