POJ 3264 Balanced Lineup

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转载最新推荐文章于 2023-10-07 08:26:33 发布 · 54 阅读

本文介绍了一种解决区间最值查询问题的有效算法——RMQ算法。该算法通过预处理实现了快速查询区间内的最大值和最小值，进而求得区间内元素的最大差值。适用于奶牛高度差查询等区间查询问题。

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Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 27620		Accepted: 12987
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

Sample Output

6
3
0

Source

USACO 2007 January Silver
RMQ算法模版题。注意运算符的优先级 1 + 1<<2 和1 + （1<<2）的结果是不一样的
同样1<<2 + 1 和(1<<2)+1的结果也是不一样的。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#define N 51000
using namespace std;
int Max[N][20],Min[N][20],a[N],n,m;
int main()
{
    //freopen("data1.in","r",stdin);
    void RMQ_init();
    void RMQ(int l,int r,int &res1,int &res2);
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        RMQ_init();
        for(int i=1;i<=m;i++)
        {
            int x,y;
            scanf("%d %d",&x,&y);
            int res1,res2;
            RMQ(x,y,res1,res2);
            printf("%d\n",res2 - res1);
        }
    }
    return 0;
}
void RMQ_init()
{
    for(int i=1;i<=n;i++)
    {
        Max[i][0] = a[i];
        Min[i][0] = a[i];
    }
    for(int j=1;(1<<j)<=n;j++)
    {
        for(int i=1;i +(1<<(j-1))<=n;i++)
        {
            Min[i][j] = min(Min[i][j-1],Min[i +(1<<(j-1))][j-1]);
            Max[i][j] = max(Max[i][j-1],Max[i +(1<<(j-1))][j-1]);
        }
    }
}
void RMQ(int l,int r,int &res1,int &res2)
{
    int k = 0;
    while(1<<(k+1)<=(r - l + 1))
    {
        k++;
    }
    res1 = min(Min[l][k],Min[r -(1<<k)+1][k]);
    res2 = max(Max[l][k],Max[r -(1<<k)+1][k]);
}