POJ 3264 Balanced Lineup(簡單的RMQ)

畫船聽雨發表於2014-02-26
MQ

話說剛開始學習線段樹的時候就拿這題試了一下水，給過了。現在學習RMQ再過一遍，感覺這種方式即好寫又快啊，不錯啊。演算法就不解釋了啊。大家可以看一下劉汝佳寫的大白書。感覺寫的挺好的。

RMQ第一題。

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

6 3
1
7
3
4
2
5
1 5
4 6
2 2

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;
}


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