﻿ P2880 [USACO07JAN]平衡的阵容Balanced Lineup - 鸿网互联

# P2880 [USACO07JAN]平衡的阵容Balanced Lineup

## 题目描述

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.

## 输入输出格式

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.

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.

## 输入输出样例

```6 3
1
7
3
4
2
5
1 5
4 6
2 2```

```6
3
0裸地ST表```
``` 1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<cmath>
5 #include<algorithm>
6 #define LL long long
7 #define lb(x)    ((x)&(-x))
8 using namespace std;
9 const int MAXN=500001;
11 {
12     char c=getchar();int x=0,f=1;
13     while(c<'0'||c>'9')    {if(c=='-')    f=-1;c=getchar();}
14     while(c>='0'&&c<='9')    x=x*10+c-48,c=getchar();return x*f;
15 }
16 int dpmin[MAXN][21];
17 int dpmax[MAXN][21];
18 int n,m;
19 int getmax(int l,int r)
20 {
21     int k=log2(r-l+1);
22     return max(dpmax[l][k],dpmax[r-(1<<k)+1][k]);
23 }
24 int getmin(int l,int r)
25 {
26     int k=log2(r-l+1);
27     return min(dpmin[l][k],dpmin[r-(1<<k)+1][k]);
28 }
29 int main()
30 {
32     for(int i=1;i<=n;i++)
34     for(int i=1;i<=19;i++)
35         for(int j=1;j<=n&&j+(1<<i)-1<=n;j++)
36             dpmax[j][i]=max(dpmax[j][i-1],dpmax[j+(1<<i-1)][i-1]),
37             dpmin[j][i]=min(dpmin[j][i-1],dpmin[j+(1<<i-1)][i-1]);
38     for(int i=1;i<=m;i++)
39     {