POJ 3264 Balanced Lineup(简单的RMQ)

最新推荐文章于 2023-10-07 08:26:33 发布

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最新推荐文章于 2023-10-07 08:26:33 发布

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分类专栏： POJ刷题数据结构 POJ练习文章标签： poj

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本文介绍如何使用线段树和RMQ算法快速解决特定问题，包括问题背景、算法解释、实例演示及代码实现。通过实际案例展示算法的实用性和效率。

话说刚开始学习线段树的时候就拿这题试了一下水，给过了。现在学习RMQ再过一遍，感觉这种方式即好写又快啊，不错啊。算法就不解释了啊。大家可以看一下刘汝佳写的大白书。感觉写的挺好的。

RMQ第一题。

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 31261		Accepted: 14730
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

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-7
#define M 10001000
#define LL __int64
//#define LL long long
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
const int maxn = 505000;

using namespace std;

int Dp_Max[maxn][20];
int Dp_Min[maxn][20];
int num[maxn];

int main()
{
    int n, m;
    while(~scanf("%d %d",&n, &m))
    {
        for(int i = 1; i  <= n; i++)
            scanf("%d",&num[i]);
        for(int i = 1; i <= n; i++)
        {
            Dp_Max[i][0] = num[i];
            Dp_Min[i][0] = num[i];
        }
        for(int j = 1; (1<<j) <= n; j++)
        {
            for(int i = 1; i+j-1<= n; i++)
            {
                Dp_Max[i][j] = max(Dp_Max[i][j-1], Dp_Max[i+(1<<(j-1))][j-1]);
                Dp_Min[i][j] = min(Dp_Min[i][j-1], Dp_Min[i+(1<<(j-1))][j-1]);
            }
        }
        int Max, Min;
        while(m--)
        {
            int l, r;
            int k = 0;
            scanf("%d %d",&l, &r);
            while(1<<(k+1) <= r-l+1)
                k++;
            Max = max(Dp_Max[l][k], Dp_Max[r-(1<<k)+1][k]);
            Min = min(Dp_Min[l][k], Dp_Min[r-(1<<k)+1][k]);
            printf("%d\n",Max-Min);
        }
    }
    return 0;
}