POJ

RMQ (Range Minimum/Maximum Query)问题是指：对于长度为n的数列A，回答若干询问RMQ(A,i,j)(i,j<=n)，返回数列A中下标在i,j里的最小(大）值，也就是说。RMQ问题是指求区间最值的问题。

Balanced Lineup

Time Limit: 5000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

Submit Status

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

1.朴素（遍历）： 复杂度O(n)-O(qn)。

2.线段树 ：复杂度O(n)-O(qlogn)。

3.ST（Sparse Table）算法 ：O(nlogn)-O(q)

说下ST算法，由于每一个查询仅仅有O(1)。在处理大量查询的时候有优势。

<1>.预处理（动态规划DP）

对A[i]数列，F[i][j] 表示从第i个数起连续2^j 中的最大值(DP的状态)，能够看到，F[i][0] 表示的是A[i]（DP的初始值）。

最后。状态转移方程是

F[i][j]=max(F[i][j-1],F[i+2^(j-1)][j-1])

<2>查询

若查询区间为（a。b），区间长度为b-a+1，取k=log2(b-a+1)，则Max（a。b）=max（F[a][k]。F[b-2^k+1][k]）。

1.ST算法

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#define max(a,b) (a>b?a:b)  
#define min(a,b) (a<b?a:b)
using namespace std;

const int MAXN = 50050;

int mins[MAXN][20];
int maxs[MAXN][20];

void RMQ(int n)
{
	for (int j = 1; (1 << j) <= n;j++)
	for (int i = 1; i + (1 << j) - 1 <= n; i++)
	{
		int p = (1 << (j - 1));
		mins[i][j] = min(mins[i][j - 1], mins[i + p][j - 1]);
		maxs[i][j] = max(maxs[i][j - 1], maxs[i + p][j - 1]);
	}
}

int queryMin(int l, int r)
{
	int k = log((double)(r - l + 1))/log(2.0);
	return min(mins[l][k], mins[r - (1 << k) + 1][k]);
}

int queryMax(int l, int r)
{
	int k = log((double)(r - l + 1))/log(2.0);
	return max(maxs[l][k], maxs[r - (1 << k) + 1][k]);
}

int main()
{
	int n, q;
	scanf("%d%d", &n, &q);
	int num;
	for (int i = 1; i <= n; i++)
	{
		scanf("%d", &num);
		maxs[i][0] = mins[i][0] = num;
	}
	RMQ(n);
	int a, b;
	int ans;
	for (int i = 0; i < q; i++)
	{
		scanf("%d%d", &a, &b);
		ans= queryMax(a, b) - queryMin(a, b);
		printf("%d
", ans);
	}
}

2.线段树

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)

using namespace std;

const int MAXN = 50050;

int num[MAXN];

struct node
{
	int r;
	int l;
	int Max;
	int Min;
}tree[3*MAXN];

void build(int l, int r, int i)
{
	tree[i].l = l; tree[i].r = r;
	if (l == r)
	{
		tree[i].Max = tree[i].Min = num[l];
		return;
	}
	int m = (l + r) >> 1, ls = i << 1, rs = ls + 1;
	build(l, m, ls);
	build(m + 1, r, rs);
	tree[i].Max = max(tree[rs].Max, tree[ls].Max);
	tree[i].Min = min(tree[rs].Min, tree[ls].Min);
}

int queryMax(int l, int r, int i)
{
	if (tree[i].l == l&&tree[i].r == r)
		return tree[i].Max;
	int m = (tree[i].l + tree[i].r) >> 1, ls = i << 1, rs = ls + 1;
	if (r <= m) return queryMax(l, r, ls);
	else if (l > m) return queryMax(l, r, rs);
	else return max(queryMax(l, m, ls), queryMax(m + 1, r, rs));
}

int queryMin(int l, int r, int i)
{
	if (tree[i].l == l&&tree[i].r == r)
		return tree[i].Min;
	int m = (tree[i].l + tree[i].r) >> 1, ls = i << 1, rs = ls + 1;
	if (r <= m) return queryMin(l, r, ls);
	else if (l > m) return queryMin(l, r, rs);
	else return min(queryMin(l, m, ls), queryMin(m + 1, r, rs));
}

int  main()
{
	int n, q;
	scanf("%d%d", &n, &q);
	for (int i = 1; i <= n; i++)
		scanf("%d", &num[i]);
	build(1, n, 1);
	int a, b;
	int ans;
	for (int i = 0; i < q; i++)
	{
		scanf("%d%d", &a, &b);
		ans = queryMax(a, b, 1) - queryMin(a, b, 1);
		printf("%d
", ans);
	}
}

參考了http://blog.csdn.net/niushuai666/article/details/6624672/