zoukankan      html  css  js  c++  java
  • POJ 3264 区间最大最小值Sparse_Table算法

    题目链接:http://poj.org/problem?id=3264

     Balanced Lineup

    Time Limit: 5000MS   Memory Limit: 65536K
    Total Submissions: 47515   Accepted: 22314
    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

    题目大意:给定一个数组,然后给一个区间[a,b],输出从a到b中的最大值和最小值的差。
    RMQ 和 线段树 均可
    RMQ代码:
     1 #include <stdio.h>
     2 #define MAX(a,b) (a>b ? a:b)
     3 #define MIN(a,b) (a>b ? b:a)
     4 #define N 50010
     5 int a[N],ma[N][25],mi[N][25];
     6 
     7 void ST(int n)
     8 {
     9    int i,j;
    10    for(i=1;i<=n;i++) 
    11        mi[i][0]=ma[i][0]=a[i];
    12        
    13     for (j = 1; (1<<j) <= n; j ++)
    14     for (i = 1; i + (1<<j)-1 <= n; i ++)
    15     {
    16         ma[i][j]=MAX(ma[i][j-1],ma[i+(1<<(j-1))][j-1]);
    17         mi[i][j]=MIN(mi[i][j-1],mi[i+(1<<(j-1))][j-1]);
    18     }
    19 }
    20 
    21 int rmq(int a,int b)
    22 {
    23     int k = 0;
    24     while((1<<(k+1)) <= b-a+1) k++;
    25     return MAX(ma[a][k],ma[b-(1<<k)+1][k])-MIN(mi[a][k],mi[b-(1<<k)+1][k]);
    26 }
    27 
    28 int main()
    29 {
    30    int n,i,q,x,y;
    31    while(scanf("%d %d",&n,&q)!=-1)
    32    {
    33        for(i=1;i<=n;i++)
    34            scanf("%d",a+i);
    35        ST(n);
    36        for(i=1;i<=q;i++)
    37        {
    38            scanf("%d %d",&x,&y);
    39            printf("%d
    ",rmq(x,y));
    40        }
    41    }
    42    return 0;
    43 }
    View Code
     
  • 相关阅读:
    sop服务治理
    调用链跟踪基本原理
    spring源码之bean的初始化及循环引用
    单调栈
    SpringBoot启动源码及自定义starter
    深入理解kafka
    Netty小结
    Oooooooooooooooooooooooooooooooooooooooooops, AN EMPTY AREA here.
    牛客OI周赛7-提高组 A 小睿睿的等式
    素数筛法求素数
  • 原文地址:https://www.cnblogs.com/yoke/p/5858054.html
Copyright © 2011-2022 走看看