zoukankan      html  css  js  c++  java
  • poj--3264 Balanced Lineup(裸的RMQ)

    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.. NQ+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
    题意:给你一个数列,不停地询问你区间【L,R】的最大值和最小值之差。
    思路:由于数据特别大,所以暴力的话一定会超时的。所以考虑到用 RMQ ,经过简单的学习就能把这个题目给A了。
    RMQ博客:http://blog.csdn.net/liang5630/article/details/7917702

    AC代码:
     1 #include <iostream>
     2 #include<cstdio>
     3 using namespace std;
     4 int a[200005],b[200005][40],c[200005][40];
     5 int main()
     6 {
     7     int n,m,l,r;
     8     while(~scanf("%d%d",&n,&m))
     9     {
    10         for(int i=1; i<=n; i++)
    11         {
    12             scanf("%d",&a[i]);
    13             b[i][0]=c[i][0]=a[i];
    14         }
    15         for(int j=1; (1<<j)<=n; j++)
    16             for(int i=1; i+(1<<(j-1))<=n; i++)
    17             {
    18                 b[i][j]=max(b[i][j-1],b[i+(1<<(j-1))][j-1]);
    19                 c[i][j]=min(c[i][j-1],c[i+(1<<(j-1))][j-1]);
    20                 //printf("%d %d %d %d
    ",i,j,b[i][j],c[i][j]);
    21             }
    22         for(int i=0; i<m; i++)
    23         {
    24             scanf("%d%d",&l,&r);
    25             int k=0;
    26             while((1<<k)<=r-l+1)
    27                 k++;
    28             printf("%d
    ",max(b[l][k-1],b[r-(1<<(k-1))+1][k-1])-min(c[l][k-1],c[r-(1<<(k-1))+1][k-1]));
    29         }
    30     }
    31     return 0;
    32 }
    View Code
     
  • 相关阅读:
    Combox小问题
    数据库登录名,用户,角色以及权限分配
    UDP初识
    AJax 无刷新构建DataTable
    批量修改数据库构架SQL
    Jquery Ajax
    Linq中使用Group By
    对象的消息模型
    P2P网络技术概览与实现原理
    ajax(1)
  • 原文地址:https://www.cnblogs.com/wang-ya-wei/p/5988424.html
Copyright © 2011-2022 走看看