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?Binclusive.

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

RMQ（Range Minimum/Maximum Query），即區間最值查詢

RMQ是指這樣一個問題：對于長度為n的數列A，回答若干次詢問RMQ(i,j)，返回數列A中下標在區間[i,j]中的最小/大值。

根據詢問，修改，單點，區間，在線，離線的各種組合大概有8類問題

一般就是4種解決辦法吧。

1、線段樹

2、樹狀數組

3、ST表

4、差分

本文介紹一種比較高效的ST算法解決這個問題。

ST（Sparse Table）算法可以在O(nlogn)時間內進行預處理，然后在O(1)時間內回答每個查詢。

1）預處理

設A[i]是要求區間最值的數列，F[i, j]表示從第i個數起連續2^j個數中的最大值。（DP的狀態）

例如：

A數列為：3 2 4 5 6 8 1 2 9 7

F[1，0]表示第1個數起，長度為2^0=1的最大值，其實就是3這個數。

同理 F[1,1] = max(3,2) = 3, F[1，2]=max(3,2,4,5) = 5，F[1，3] = max(3,2,4,5,6,8,1,2) = 8;

并且我們可以容易的看出F[i,0]就等于A[i]。（DP的初始值）

我們把F[i，j]平均分成兩段（因為F[i，j]一定是偶數個數字），從 i 到i + 2 ^ (j - 1) - 1為一段，i + 2 ^ (j - 1)到i + 2 ^ j - 1為一段(長度都為2 ^ (j - 1))。

于是我們得到了狀態轉移方程：

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

2)查詢

假如我們需要查詢的區間為(i,j)，那么我們需要找到覆蓋這個閉區間?(左邊界取i，右邊界取j)?的最小冪（可以重復，比如查詢1，2，3，4，5，我們可以查詢1234和2345）。

因為這個區間的長度為j - i + 1,所以我們可以取k=log2( j - i + 1)，則有：RMQ(i, j)=max{F[i , k], F[ j - 2 ^ k + 1, k]}。

舉例說明，要求區間[1，5]的最大值，k = log2（5 - 1 + 1）= 2，即求?max(F[1, 2]，F[5 - 2 ^ 2 + 1, 2])=max(F[1, 2]，F[2, 2])；

void ST(int n) {for (int i = 1; i <= n; i++)dp[i][0] = A[i];for (int j = 1; (1 << j) <= n; j++) {for (int i = 1; i + (1 << j) - 1 <= n; i++) {dp[i][j] = max(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);}} } int RMQ(int l, int r) {int k = 0;while ((1 << (k + 1)) <= r - l + 1) k++;return max(dp[l][k], dp[r - (1 << k) + 1][k]); }

poj3264

RMQ問題 ST算法

/* poj3264 Balanced Lineup */ #include <cstdio> #include <cmath> #include<algorithm> #define MAXN (50000 + 10) using namespace std; int cows[MAXN]; int st_max[MAXN][20]; int st_min[MAXN][20]; int n, q, l, r; void initst_max() { for (int i = 0; i < n; i++) st_max[i][0] = cows[i]; for (int j = 1; (1<<j) < n; ++j) { for (int i = 0; i + (1<<j) <= n; ++i) { st_max[i][j] = max(st_max[i][j-1], st_max[i+(1<<(j-1))][j-1]); } } } void initst_min() { for (int i = 0; i < n; i++) st_min[i][0] = cows[i]; for (int j = 1; (1<<j) < n; ++j) { for (int i = 0; i + (1<<j) <= n; ++i) { st_min[i][j] = min(st_min[i][j-1], st_min[i+(1<<(j-1))][j-1]); } } } int queryst_max(int l, int r) { int k = (int)(log(r-l+1.0)/log(2.0)); return max(st_max[l][k], st_max[r-(1<<k)+1][k]); } int queryst_min(int l, int r) { int k = (int)(log(r-l+1.0)/log(2.0)); return min(st_min[l][k], st_min[r-(1<<k)+1][k]); } int main() { while (scanf("%d %d", &n, &q) != EOF) { for (int i = 0; i < n; ++i) scanf("%d", &cows[i]); initst_max(); initst_min(); while (q--) { scanf("%d %d", &l, &r); if (l == r) printf("0\n"); else { int a = queryst_max(l-1, r-1); int b = queryst_min(l-1, r-1); printf("%d\n", a - b); } } } return 0; }

線段樹C++

#include <iostream> #include <cstdio> using namespace std; #define INF 0x3f3f3f3f const int maxn = (5e4+10);struct node{int l,r,max,min; };int a[maxn]; node tree[maxn<<2];static int max(int a,int b) {return a>b?a:b; }static int min(int a,int b) {return a<b?a:b; }static void pushup(int rt) {tree[rt].max=max(tree[rt<<1].max, tree[rt<<1|1].max);tree[rt].min=min(tree[rt<<1].min, tree[rt<<1|1].min); } static void build(int rt,int l,int r) {tree[rt].l=l;tree[rt].r=r;if(l==r) {tree[rt].max=tree[rt].min=a[l];return;}int mid=(l+r)>>1;build(rt<<1,l,mid);build(rt<<1|1,mid+1,r);pushup(rt); } static int queryMax(int rt,int l,int r) {if(tree[rt].l==l&&tree[rt].r==r) {return tree[rt].max;}int mid=(tree[rt].l+tree[rt].r)>>1;if(r<=mid)return queryMax(rt<<1,l,r);else if(l>mid)return queryMax(rt<<1|1,l,r);elsereturn max(queryMax(rt<<1,l,mid),queryMax(rt<<1|1,mid+1,r)); } static int queryMin(int rt,int l,int r) {if(tree[rt].l==l&&tree[rt].r==r) {return tree[rt].min;}int mid=(tree[rt].l+tree[rt].r)>>1;if(r<=mid)return queryMin(rt<<1,l,r);else if(l>mid)return queryMin(rt<<1|1,l,r);elsereturn min(queryMin(rt<<1,l,mid),queryMin(rt<<1|1,mid+1,r)); }int main() { int q,n;while(scanf("%d%d",&n,&q)!=EOF) {for(int i=1;i<=n;i++)scanf("%d",&a[i]);build(1,1,n);while((q--)!=0) {int l;int r;scanf("%d%d",&l,&r);printf("%d\n",queryMax(1,l,r)-queryMin(1,l,r));}}}

線段樹單點查詢，Java

import java.util.*; import java.math.*; class node{int l,r,max,min;node(int l,int r){this.l=0;this.r=0;this.max=0;this.min=0;} }public class Main{static int maxn=(int)(5e4+10);static int[] a=new int[maxn];static node[] tree=new node[maxn<<2];static void init() {for(int i=0;i<(maxn<<2);i++) {tree[i]=new node(0,0);}}static int max(int a,int b) {return a>b?a:b;}static int min(int a,int b) {return a<b?a:b;}static void pushup(int rt) {tree[rt].max=max(tree[rt<<1].max, tree[rt<<1|1].max);tree[rt].min=min(tree[rt<<1].min, tree[rt<<1|1].min);}static void build(int rt,int l,int r) {tree[rt].l=l;tree[rt].r=r;if(l==r) {tree[rt].max=tree[rt].min=a[l];return;}int mid=(l+r)>>1;build(rt<<1,l,mid);build(rt<<1|1,mid+1,r);pushup(rt);}static int queryMax(int rt,int l,int r) {if(tree[rt].l==l&&tree[rt].r==r) {return tree[rt].max;}int mid=(tree[rt].l+tree[rt].r)>>1;if(r<=mid)return queryMax(rt<<1,l,r);else if(l>mid)return queryMax(rt<<1|1,l,r);elsereturn max(queryMax(rt<<1,l,mid),queryMax(rt<<1|1,mid+1,r));}static int queryMin(int rt,int l,int r) {if(tree[rt].l==l&&tree[rt].r==r) {return tree[rt].min;}int mid=(tree[rt].l+tree[rt].r)>>1;if(r<=mid)return queryMin(rt<<1,l,r);else if(l>mid)return queryMin(rt<<1|1,l,r);elsereturn min(queryMin(rt<<1,l,mid),queryMin(rt<<1|1,mid+1,r));}public static void main(String[] args) {Scanner cin=new Scanner(System.in);init();while(cin.hasNext()) {int n=cin.nextInt();int q=cin.nextInt();for(int i=1;i<=n;i++)a[i]=cin.nextInt();build(1,1,n);while((q--)!=0) {int l=cin.nextInt();int r=cin.nextInt();System.out.println(queryMax(1,l,r)-queryMin(1,l,r));}}cin.close();} }

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