zoukankan      html  css  js  c++  java
  • Stars(树状数组或线段树)

    Stars
    Time Limit: 1000MS Memory Limit: 65536K
    Total Submissions: 37323 Accepted: 16278

    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

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

    Source
    Ural Collegiate Programming Contest 1999
    线段树和树状数组皆可AC
    树状数组(和金巨学的)

    #include <iostream>
    #include <cmath>
    #include <cstring>
    #include <cstdlib>
    #include <stdio.h>
    #include <string>
    #include <queue>
    #include <vector>
    #include <algorithm>
    #define LL long long
    using namespace std;
    
    const int INF = 0x3f3f3f3f;
    
    const int MAX = 32010;
    
    int c[MAX];
    
    int level[MAX/2];
    
    int lowbit(int x)
    {
        return x&(-x);
    }
    
    int sum(int x)
    {
        int sum=0;
        while(x>0)
        {
            sum+=c[x];
            x-=lowbit(x);
        }
        return sum;
    }
    
    void add(int x)
    {
        while(x<MAX)
        {
            c[x]++;
            x+=lowbit(x);
        }
    }
    
    int main()
    {
        int n;
        int x,y;
        memset(level,0,sizeof(level));
        memset(c,0,sizeof(c));
        scanf("%d",&n);
        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
    ",level[i]);
        }
        return 0;
    }
    

    线段树

    #include <iostream>
    #include <cmath>
    #include <cstring>
    #include <cstdlib>
    #include <stdio.h>
    #include <string>
    #include <queue>
    #include <vector>
    #include <algorithm>
    #define LL long long
    using namespace std;
    
    const int INF = 0x3f3f3f3f;
    
    const LL MAX = 32000;
    
    int a[MAX*4];
    int level[MAX];
    
    
    int Query(int L,int R,int l,int r,int site)
    {
        if(L==l&&R==r)
        {
            return a[site];
        }
        int mid=(L+R)>>1;
        if(r<=mid)
        {
            return Query(L,mid,l,r,site<<1);
        }
        else if(l>mid)
        {
            return Query(mid+1,R,l,r,site<<1|1);
        }
        else
        {
            return Query(L,mid,l,mid,site<<1)+Query(mid+1,R,mid+1,r,site<<1|1);
        }
    }
    void update(int L,int R,int s,int site)
    {
        a[site]++;
        if(L==R)
        {
            return ;
        }
        int mid=(L+R)>>1;
        if(s<=mid)
        {
            update(L,mid,s,site<<1);
        }
        else
        {
            update(mid+1,R,s,site<<1|1);
        }
    }
    int main()
    {
        int n;
        int x,y;
        memset(level,0,sizeof(level));
        memset(a,0,sizeof(a));
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d %d",&x,&y);
            level[Query(0,MAX,0,x,1)]++;
            update(0,MAX,x,1);
        }
        for(int i=0;i<n;i++)
        {
            printf("%d
    ",level[i]);
        }
        return 0;
    }
    
  • 相关阅读:
    关于asp.net中Repeater控件的一些应用
    Linux查看程序端口占用情况
    php 验证身份证有效性,根据国家标准GB 11643-1999 15位和18位通用
    给Nginx配置一个自签名的SSL证书
    让你提升命令行效率的 Bash 快捷键 [完整版]
    关系数据库常用SQL语句语法大全
    php 跨域 form提交 2种方法
    Vimium~让您的Chrome起飞
    vim tab设置为4个空格
    CENTOS 搭建SVN服务器(附自动部署到远程WEB)
  • 原文地址:https://www.cnblogs.com/juechen/p/5256001.html
Copyright © 2011-2022 走看看