zoukankan      html  css  js  c++  java
  • HDU 1541 Stars(树状数组)

    Stars

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 6479    Accepted Submission(s): 2570


    Problem 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
    5 1 1 5 1 7 1 3 3 5 5
     

    Sample Output
    1 2 1 1 0
     

    Source

    Ural Collegiate Programming Contest 1999


    点击打开链接


    #include<iostream>
    #include<algorithm>
    #include<stdio.h>
    #include<string.h>
    #include<stdlib.h>
    
    #define N 32001
    
    using namespace std;
    
    int n;
    int c[N];
    int num[N];
    
    int lowbit(int x) {
        return x&(-x);
    }
    
    int getsum(int x) {
        int s = 0;
        while(x>0) {
            s += c[x];
            x -= lowbit(x);
        }
        return s;
    }
    
    void build(int x,int y) {
        while(x<=N) {
            c[x] += y;
            x += lowbit(x);
        }
    }
    
    int main() {
        while(scanf("%d",&n)!=EOF) {
            memset(num,0,sizeof(num));
            memset(c,0,sizeof(c));
            int x,y;
            for(int i=0; i<n; i++) {
                scanf("%d%d",&x,&y);
                num[getsum(x+1)]++ ;
                build(x+1,1);
            }
            for(int i=0; i<n; i++) {
                printf("%d
    ",num[i]);
            }
        }
        return 0;
    }
    


  • 相关阅读:
    忠告20岁的年轻人
    mac电脑好用的工具总结
    idea 配置
    mac 安装mysql5.7.28附安装包
    国内外优秀网站收集
    MySql 数据库、数据表操作
    Java 高效代码50例
    Mac 修改版本号
    sql 语句系列(删库跑路系列)[八百章之第七章]
    sql 语句系列(更新系列)[八百章之第六章]
  • 原文地址:https://www.cnblogs.com/tlnshuju/p/7347213.html
Copyright © 2011-2022 走看看