POJ 3264 Balanced Lineup (ST算法)

题目链接：POJ 3264

Problem Description

For the daily milking, Farmer John’s $N$ cows $(1 \le N \le 50,000)$ 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 $Q (1 \le Q \le 200,000)$ potential groups of cows and their $heights (1 \le height \le 1,000,000)$. 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

Sample Output

1
2
3

6
3
0

Source

USACO 2007 January Silver

Solution

题意

给定 $n$ 个数，$q$ 个询问。每个询问包含 $l$ 和 $r$，求区间 $[l, r]$ 内的最大值减最小值的差值。

题解

ST算法

ST 算法模板题。

Code

#include <cstdio>
#include <iostream>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn = 1e5 + 10;

int a[maxn];
int f[maxn][50], g[maxn][50];
int n, q;

void st_prework() {
    for(int i = 1; i <= n; ++i) {
        f[i][0] = a[i];
        g[i][0] = a[i];
    }
    int t = log(n) / log(2) + 1;
    for(int j = 1; j < t; ++j) {
        for(int i = 1; i <= n - (1 << j) + 1; ++i) {
            f[i][j] = max(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);
            g[i][j] = min(g[i][j - 1], g[i + (1 << (j - 1))][j - 1]);
        }
    }
}

int st_query(int l, int r) {
    int k = log(r - l + 1) / log(2);
    return max(f[l][k], f[r - (1 << k) + 1][k]) - min(g[l][k], g[r - (1 << k) + 1][k]);
}

int main() {
    scanf("%d%d", &n, &q);
    for(int i = 1; i <= n; ++i) {
        scanf("%d", &a[i]);
    }
    st_prework();
    for(int i = 0; i < q; ++i) {
        int l, r;
        scanf("%d%d", &l, &r);
        printf("%d\n", st_query(l, r));
    }
    return 0;
}