zoukankan      html  css  js  c++  java
  • poj3264Balanced Lineup(倍增ST表)

    Balanced Lineup
    Time Limit: 5000MS   Memory Limit: 65536K
    Total Submissions: 52328   Accepted: 24551
    Case Time Limit: 2000MS

    Description

    For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things simple, he will take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to have fun they should not differ too much in height.

    Farmer John has made a list of Q (1 ≤ Q ≤ 200,000) potential groups of cows and their heights (1 ≤ height ≤ 1,000,000). For each group, he wants your help to determine the difference in height between the shortest and the tallest cow in the group.

    Input

    Line 1: Two space-separated integers, N and Q
    Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i 
    Lines N+2..N+Q+1: Two integers A and B (1 ≤ A ≤ B ≤ N), representing the range of cows from A to B inclusive.

    Output

    Lines 1..Q: Each line contains a single integer that is a response to a reply and indicates the difference in height between the tallest and shortest cow in the range.

    Sample Input

    6 3
    1
    7
    3
    4
    2
    5
    1 5
    4 6
    2 2

    Sample Output

    6
    3
    0

    //先敲个板子
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    
    #define maxn 1000000
    
    using namespace std;
    int n,m,ans,x,y,a[maxn],p[maxn];
    int f1[maxn][25],f2[maxn][25];
    
    inline int init()
    {
        int x=0,f=1;char c=getchar();
        while(c>'9'||c<'0'){if(c=='-')f=-1;c=getchar();}
        while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
        return x*f;
    }
    
    int max(int x,int y){return x>y?x:y;}
    int min(int x,int y){return x<y?x:y;}
    
    void ST()
    {
        for(int i=1;i<=n;i++)
          f1[i][0]=f2[i][0]=a[i];
        for(int j=1;j<=20;j++)
          for(int i=1;i+(1<<j)-1<=n;i++)
            {
                f1[i][j]=min(f1[i][j-1],f1[i+(1<<j-1)][j-1]);
                f2[i][j]=max(f2[i][j-1],f2[i+(1<<j-1)][j-1]);
            }
        for(int i=1;i<=n;i++)
          for(int j=0;j<=20;j++)
               if((1<<j)>i)
               {
                  p[i]=j-1;
                  break;                
            }
    }
    
    int query(int l,int r)
    {
        int k=p[r-l+1];
        int ans1=min(f1[l][k],f1[r-(1<<k)+1][k]);
        int ans2=max(f2[l][k],f2[r-(1<<k)+1][k]);
        return ans2-ans1;
    }
    
    int main()
    {
        n=init();m=init();
        for(int i=1;i<=n;i++)
          a[i]=init();
        ST();
        for(int i=1;i<=m;i++)
        {
            x=init();y=init();
            printf("%d
    ",query(x,y));
        }
        return 0;
    }
    折花枝,恨花枝,准拟花开人共卮,开时人去时。 怕相思,已相思,轮到相思没处辞,眉间露一丝。
  • 相关阅读:
    [JSOI2010]满汉全席 2sat
    (转)MongoDB实战开发 【零基础学习,附完整Asp.net示例】
    (转)ASP.NET的Cookie跨域问题
    (转)发一个自己写的账号管理软件
    (转)Silverlight学习点滴之一——使用WCF RIA构建应用
    (转)再议依赖注入
    (转)【探索发现】winform 网络传输时候封包与解包心得
    (转)使用Entity Framework和WCF Ria Services开发SilverLight之1:简单模型
    (转)LINQ to Entities 多条件动态查询
    (转)最老程序员创业札记:全文检索、数据挖掘、推荐引擎应用15
  • 原文地址:https://www.cnblogs.com/L-Memory/p/6815944.html
Copyright © 2011-2022 走看看