zoukankan      html  css  js  c++  java
  • poj3264

    Balanced Lineup
    Time Limit: 5000MS   Memory Limit: 65536K
    Total Submissions: 45777   Accepted: 21499
    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

    Source

    题解:
    求区间最大值-最小值
     
    RMQ AC代码
    #include<cstdio>
    #include<cmath>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    #define N 50010
    int n,m,a[N],f[N][25],g[N][25];
    inline int read(){
        register int x=0,f=1;
        register char ch=getchar();
        while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
        while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
        return x*f;
    }
    inline void RMQ(){
        for(int j=1;j<=20;j++){
            for(int i=1;i+(1<<j)-1<=n;i++){
                f[i][j]=max(f[i][j-1],f[i+(1<<j-1)][j-1]),
                g[i][j]=min(g[i][j-1],g[i+(1<<j-1)][j-1]);
            }
        }
    }
    inline int query(int i,int j){
        int k=log(j-i+1)/log(2);
        return max(f[i][k],f[j-(1<<k)+1][k])-min(g[i][k],g[j-(1<<k)+1][k]);
    }
    int main(){
        n=read();m=read();
        for(int i=1;i<=n;i++) g[i][0]=f[i][0]=read();
        RMQ();
        for(int i=1,l,r;i<=m;i++) l=read(),r=read(),printf("%d
    ",query(l,r));
        return 0;
    }

    线段树代码,自己写吧。

  • 相关阅读:
    stm32f103 SPI单线TX发数据来驱动LCD
    【转】常见排序算法
    [转]命令行 Subversion 入门
    JLINK V8 Keil MDK4.10 STM32
    字符串表示的大整数相加
    字符串反转
    字符串表示的大整数相乘
    猴子选大王
    [转]Posix-- 互斥锁 条件变量 信号量
    [转]Openwrt的Inittab
  • 原文地址:https://www.cnblogs.com/shenben/p/5705220.html
Copyright © 2011-2022 走看看