POJ 3264 Balanced Lineup (RMQ模板)

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本文介绍了一种解决区间最大值和最小值问题的有效算法。通过使用预处理技术和动态规划思想，该算法能在对数时间内查询任意指定区间的最大值和最小值，适用于处理大量查询请求。文章详细解释了算法原理，提供了完整的C++代码实现，并通过USACO比赛题目进行实例演示。

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 65283		Accepted: 30409
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

题目链接：http://poj.org/problem?id=3264

题目大意：求区间最大值减最小值

题目分析：RMQ模板题

#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
int const MAX = 50005;
int n, q, l, r, a[MAX], ma[MAX][21], mi[MAX][21];

void Init_RMQ() {
    for (int i = 1; i <= n; i++) {
        ma[i][0] = mi[i][0] = a[i];
    }
    for (int j = 1; j <= 20; j++) {
        for (int i = 1; i + (1 << j) - 1 <= n; i++) {
            ma[i][j] = max(ma[i][j - 1], ma[i + (1 << (j - 1))][j - 1]);
            mi[i][j] = min(mi[i][j - 1], mi[i + (1 << (j - 1))][j - 1]);
        }
    }
}

int solve(int l, int r) {
    int k = (int)(log((double)(r - l + 1)) / log(2.0));
    return max(ma[l][k], ma[r - (1 << k) + 1][k]) - min(mi[l][k], mi[r - (1 << k) + 1][k]);
}

int main() {
    scanf("%d %d", &n, &q);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
    }
    Init_RMQ();
    while (q--) {
        scanf("%d %d", &l, &r);
        printf("%d\n", solve(l, r));
    }
}