本文主要是介绍POJ-3264 Balanced Lineup(RMQ)，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
Balanced Lineup
| Time Limit: 5000MS | Memory Limit: 65536K | |
|---|---|---|
| Total Submissions: 73096 | Accepted: 33626 | |
| 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
链接： http://poj.org/problem?id=3264
题意：
给你一个数组A，给你一段区间[l,r]，求区间最大值-最小值
思路：
裸的RMQ，直接套板子
代码：
#include <cstdio>
#include <cmath>
#include <iostream>
using namespace std;
const int maxn = 50000+10;
int a[maxn];
int maxx[maxn][17];
int minn[maxn][17];
int read()
{int num = 0;char c = getchar();while(c > '9' || c < '0')c = getchar();while(c >= '0'&& c<= '9'){num=num*10+c-'0';c = getchar();}return num;
}
void write(int x)
{if(x<0){putchar('-');x=-x;}if(x>9)write(x/10);putchar(x%10+'0');
}
int main()
{int n, q;n = read();q = read();for (int i = 0; i < n; ++i) {a[i] = read();maxx[i][0] = a[i];minn[i][0] = a[i];}for(int j = 1; (1<<j) <= n; j++) {for(int i = 0; i < n; i++) {if(i + (1<<j) - 1 < n) {maxx[i][j] = max(maxx[i][j-1], maxx[i+(1<<(j-1))][j-1]);minn[i][j] = min(minn[i][j-1], minn[i+(1<<(j-1))][j-1]);}}}int l, r;int k;for(int i = 0; i < q; i++) {l = read();r = read();k = log(r-l+1) / log(2);l--;r--;write(max(maxx[l][k], maxx[r-(1<<k)+1][k]) - min(minn[l][k], minn[r-(1<<k)+1][k]));cout << endl;}return 0;
}
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