【题解】平衡队列
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 ≤ 180,000) 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.
每天,农夫 John 的
头牛总是按同一序列排队。
有一天, John 决定让一些牛们玩一场飞盘比赛。他准备找一群在队列中位置连续的牛来进行比赛。但是为了避免水平悬殊，牛的身高不应该相差太大。John 准备了
个可能的牛的选择和所有牛的身高 hih_ihi
。他想知道每一组里面最高和最低的牛的身高差。
输入格式
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.
第一行两个数 n,q。
接下来 n 行，每行一个数
再接下来 q 行，每行两个整数 a 和 b，表示询问第 a 头牛到第 b 头牛里的最高和最低的牛的身高差。
输出格式
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.
输出共 q 行，对于每一组询问，输出每一组中最高和最低的牛的身高差。
输入输出样例
输入 #1
6 3
1
7
3
4
2
5
1 5
4 6
2 2输出 #1
6
3
0题目分析
阅读题目并提炼主干，可发现，题目是要我们求连续区域内的最值。也就是RMQ(Range Minimum/Maximum Query)区间最值问题。可以使用ST表来解决这类问题。
ST表预处理
int ST_query(int l,int r){//询问l~r区间的最值
int Lg=Log[r-l+1];//计算l~r之间长度对应的log2值
return max(stMax[l][Lg],stMax[r-(1<<Lg)+1][Lg]);
}
void ST_prework(){//ST表预处理
for(int i=1;i<=n;i++){//f[i][j]=从i开始，长为2^j区间内的最值
f[i][0]=a[i];//长为1时的最值
}
for(int j=1;j<=Log[n];j++){//根据最长区间长度log[n]进行遍历
for(int i=1;i+(1<<j)-1<=n;i++){//遍历开始位置i
f[i][j]=max(f[i][j-1],f[i+(1<<(j-1))][j-1]);//左边的最值与右边最值中较大者为整个区域的最值
}
}
}可以维护两个st表，一个放区间最大值，一个放区间最小值。
实现代码
#include <iostream>
#include <cstdio>
using namespace std;
const int N=5e4+5;
int n,q;
int h[N],f1[N][20],f2[N][20];
int Log[N];
int stMax(int l,int r){//询问l~r区间的最值
int Lg=Log[r-l+1];//计算l~r之间长度对应的log2值
return max(f1[l][Lg],f1[r-(1<<Lg)+1][Lg]);
}
int stMin(int l,int r){
int Lg=Log[r-l+1];
return min(f2[l][Lg],f2[r-(1<<Lg)+1][Lg]);
}
void ST_prework(){//ST表预处理
for(int i=1;i<=n;i++){//f[i][j]=从i开始，长为2^j区间内的最值
//f1最大 f2最小
f2[i][0]=f1[i][0]=h[i];//长为1时的最值
}
for(int j=1;j<=Log[n];j++){//根据最长区间长度log[n]进行遍历
for(int i=1;i+(1<<j)-1<=n;i++){//遍历开始位置i
f1[i][j]=max(f1[i][j-1],f1[i+(1<<(j-1))][j-1]);//左边的最值与右边最值中较大者为整个区域的最值
f2[i][j]=min(f2[i][j-1],f2[i+(1<<(j-1))][j-1]);//左边的最值与右边最值中较小者为整个区域的最值
}
}
}
int main(){
scanf("%d%d",&n,&q);
for(int i=1;i<=n;i++){
scanf("%d",&h[i]);
}
//预处理Log[]数组 ， Log[x]=log2(x)
for(int i=2;i<=n;i++){
Log[i]=Log[i/2]+1;
}
ST_prework();
int a,b;
while(q--){
scanf("%d%d",&a,&b);
printf("%d\n",stMax(a,b)-stMin(a,b));
}
return 0;
}Q.E.D.