ST算法—Balanced Lineup

[USACO07JAN] Balanced Lineup G

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 \(n(1\le n\le 5\times 10^4)\) 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 \(q(1\le q\le 1.8\times10^5)\) 个可能的牛的选择和所有牛的身高 \(h_i(1\le h_i\le 10^6,1\le i\le n)\)。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 \(h_i\)。

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。

样例 #1

样例输入 #1

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

样例输出 #1

6
3
0

分析

题目就是在查询任意区间内的最值，如果使用暴力算法时间复杂度就是O(mn)超时，这时候我们可以看到题目给定的数列是静态的，可以考虑st算法解决也就是稀疏表法，首先，我们把数列按倍增分为2的次幂个小区间，然后我们去求每个小区间的最值，我们发现每组小区间的最值可以由前一组递推过来（动态规划）最后我们把要求的区间分为两个小区间通过小区间长度len，向下取2的对数，这样就能把前后两个区间都分成我们第一步的2的次幂小区间，及时前后区间有重合部分也不影响最值，然后利用第二步直接O(1)复杂度求出最值。

代码实现

#include <bits/stdc++.h>
using namespace std;
const int N = 50005;
int n,m;
int a[N],dp_max[N][22],dp_min[N][21];
int LOG2[N];               //自己计算以2为底的对数，向下取整
void st_init(){
     LOG2[0] = -1;
     for(int i = 1;i<=N;i++)  LOG2[i]=LOG2[i>>1]+1; //不用系统log()函数，自己算
     for(int i=1;i<=n;i++){    //初始化区间长度为1时的值
         dp_min[i][0] = a[i];
         dp_max[i][0] = a[i];
     }
//int p=log2(n);                          //可倍增区间的最大次方: 2^p <= n
int p= (int)(log(double(n)) / log(2.0));  //两者写法都行
	for(int k=1;k<=p;k++)   //倍增计算小区间。先算小区间，再算大区间，逐步递推
        for(int s=1;s+(1<<k)<=n+1;s++){
            dp_max[s][k]=max(dp_max[s][k-1], dp_max[s+(1<<(k-1))][k-1]);
            dp_min[s][k]=min(dp_min[s][k-1], dp_min[s+(1<<(k-1))][k-1]);
	}
}
int st_query(int L,int R){
//  int k = log2(R-L+1);                           //3种方法求k
    int k = (int)(log(double(R-L+1)) / log(2.0));
//  int k = LOG2[R-L+1];                            //自己算LOG2
    int x = max(dp_max[L][k],dp_max[R-(1<<k)+1][k]); //区间最大
    int y = min(dp_min[L][k],dp_min[R-(1<<k)+1][k]); //区间最小
    return x-y;  //返回差值
}
int main(){
    scanf("%d%d",&n,&m);   //输入
    for(int i=1;i<=n;i++)    scanf("%d",&a[i]);
    st_init();
for(int i=1;i<=m;i++){
int L,R; scanf("%d%d",&L,&R);
printf("%d\n",st_query(L,R));
}
return 0;
}