POJ 3264 Balanced Lineup RMQ
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Balanced Lineup
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 Source
USACO 2007 January Silver
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范围最小值问题（Range Minimum Query，RMQ）。给一个 n 个元素的数组A1，A2，...，An，查询范围（L，R）
中的最小值和最大值。
下面这篇大牛的文章详细介绍了RMQ
RMQ详解
Sparse-Table算法，dp初始化，时间O(nlogn)，查询只需O(1)。
令 d(i, j) 表示从 i 开始，长度为 2^j 的一段元素的最小值，则状态转移：
d( i , j ) = min{ d( i , j - 1 ), d(i + 2^( j - 1) , j - 1 )}，原理如下图：
注意2^j <= n,因此 d 数组的元素个数不超过 nlogn ,而每一项都可以在常数时间计算完毕，故总时间为O(nlogn)。
查询时，令 k 为满足 2^k <= R - L + 1 的最大整数，则以 L 开头，以 R 结尾的两个长度为2^k 的区间合起来即覆盖了查询区间[L , R]。
代码：
#include <cstdio>
#include <vector>
#include <iostream>
using namespace std;
const int MAXN = 50000 + 10;
int num[MAXN], mind[MAXN][16], maxd[MAXN][16]; //2^16 > MAXN
int N, M;
vector<int> V;
//预处理，时间复杂度：O(nlogn)
void RMQ_init(const vector<int>& A)
{
int n = A.size();
for (int i = 0; i < n; i++) mind[i][0] = maxd[i][0] = A[i]; //初始化
for (int j = 1; (1 << j) <= n; j++) //状态转移
{
for (int i = 0; i + (1 << j) - 1 <= n; i++)
{
mind[i][j] = min(mind[i][j - 1], mind[i + (1 << (j - 1))][j - 1]);
maxd[i][j] = max(maxd[i][j - 1], maxd[i + (1 << (j - 1))][j - 1]);
}
}
}
//查询
int RMQ(int L, int R, int book)
{
int k = 0;
while ((1 << (k + 1)) <= R - L + 1) k++;
return book == 0 ? min(mind[L][k], mind[R - (1 << k) + 1][k])
: max(maxd[L][k], maxd[R - (1 << k) + 1][k]);
}
int main()
{
scanf("%d%d", &N, &M);
int x;
for (int i = 0; i < N; i++)
{
scanf("%d", &x);
V.push_back(x);
}
RMQ_init(V);
while (M--)
{
int L, R;
scanf("%d%d", &L, &R); //查询时下标从1开始，故减一
printf("%d\n", RMQ(L - 1, R - 1, 1) - RMQ(L - 1, R - 1, 0));
}
return 0;
}