zoukankan      html  css  js  c++  java
  • poj 3264 Balanced Lineup rmq

    Balanced Lineup
    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

    Source

    思路:rmq 板子题,区间最大值-区间最小值,无更新;
    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<string>
    #include<queue>
    #include<algorithm>
    #include<stack>
    #include<cstring>
    #include<vector>
    #include<list>
    #include<set>
    #include<map>
    using namespace std;
    #define ll long long
    //#define mod 1000000007
    #define pi (4*atan(1.0))
    const int N=1e5+10,M=1e6+10,inf=1e9+10;
    int a[N];
    int dpi[N][30];//存位置
    int dpa[N][30];
    int minn(int x,int y)
    {
        return a[x]<=a[y]?x:y;
    }
    void rmqi(int len)
    {
        for(int i=0; i<len; i++)
        dpi[i][0]=i;
        for(int j=1; (1<<j)<len; j++)
        for(int i=0; i+(1<<j)-1<len; i++)
        dpi[i][j]=minn(dpi[i][j-1],dpi[i+(1<<(j-1))][j-1]);
    }
    int queryi(int l,int r)
    {
        int x=(int)(log((double)(r-l+1))/log(2.0));
        return minn(dpi[l][x],dpi[r-(1<<x)+1][x]);
    }
    int maxx(int x,int y)
    {
        return a[x]>=a[y]?x:y;
    }
    void rmqa(int len)
    {
        for(int i=0; i<len; i++)
        dpa[i][0]=i;
        for(int j=1; (1<<j)<len; j++)
        for(int i=0; i+(1<<j)-1<len; i++)
        dpa[i][j]=maxx(dpa[i][j-1],dpa[i+(1<<(j-1))][j-1]);
    }
    int querya(int l,int r)
    {
        int x=(int)(log((double)(r-l+1))/log(2.0));
        return maxx(dpa[l][x],dpa[r-(1<<x)+1][x]);
    }
    int main()
    {
        int x,y,q,i,t;
        while(~scanf("%d%d",&x,&q))
        {
            for(i=0;i<x;i++)
            scanf("%d",&a[i]);
            rmqi(x);
            rmqa(x);
            while(q--)
            {
                int l,r;
                scanf("%d%d",&l,&r);
                if(l>r)
                swap(l,r);
                printf("%d
    ",a[querya(l-1,r-1)]-a[queryi(l-1,r-1)]);
            }
        }
    }
  • 相关阅读:
    微信小程序实现滚动到指定位置
    微信小程序,scroll-view组件的使用,跳转到指定的锚点/定位跳转
    小程序,报渲染层错误。图片无法渲染。
    input Input 输入判断/正则
    js,某元素在浏览器页面浮动/飘动
    前端模块化
    npx create-react-app命令不成功,更改成淘宝镜像
    深入理解jdk和jre(转)
    Java学习路线图·影响一代又一代程序员的经典书籍!(转)
    ZAB协议(转)
  • 原文地址:https://www.cnblogs.com/jhz033/p/5576797.html
Copyright © 2011-2022 走看看