NC25045 [USACO 2007 Jan S]Balanced Lineup

题目链接

题目

题目描述

For the daily milking, Farmer John's N cows (1 ≤ N ≤ 100,000) 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 ≤ 30) 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.

输入描述

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.

输出描述

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.

示例1

输入

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

输出

6
3
0

题解

方法一

知识点：线段树。

区间最值板子题，熟悉一下模板。

时间复杂度 \(O((n+q)\log n)\)

空间复杂度 \(O(n)\)

方法二

知识点：ST表。

不需要修改，因此同样也可以ST表做。

时间复杂度 \(O(n\log n + q)\)

空间复杂度 \(O(n)\)

代码

方法一

#include <bits/stdc++.h>
using namespace std;
using ll = long long;

struct T {
    int mx, mi;
    static T e() { return { (int)-2e9,(int)2e9 }; }
    friend T operator+(const T &a, const T &b) { return { max(a.mx, b.mx),min(a.mi,b.mi) }; }
};

template<class T>
class SegmentTree {
    int n;
    vector<T> node;

    T query(int rt, int l, int r, int x, int y) {
        if (r < x || y < l) return T::e();
        if (x <= l && r <= y) return node[rt];
        int mid = l + r >> 1;
        return query(rt << 1, l, mid, x, y) + query(rt << 1 | 1, mid + 1, r, x, y);
    }

public:
    SegmentTree() {}
    SegmentTree(const vector<T> &src) { init(src); }

    void init(const vector<T> &src) {
        assert(src.size());
        n = src.size() - 1;
        node.assign(n << 2, T::e());
        function<void(int, int, int)> build = [&](int rt, int l, int r) {
            if (l == r) return node[rt] = src[l], void();
            int mid = l + r >> 1;
            build(rt << 1, l, mid);
            build(rt << 1 | 1, mid + 1, r);
            node[rt] = node[rt << 1] + node[rt << 1 | 1];
        };
        build(1, 1, n);
    }

    T query(int x, int y) { return query(1, 1, n, x, y); }
};

int main() {
    std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int n, q;
    cin >> n >> q;
    vector<T> a(n + 1);
    for (int i = 1, x;i <= n;i++) cin >> x, a[i] = { x,x };
    SegmentTree<T> sgt(a);
    while (q--) {
        int l, r;
        cin >> l >> r;
        auto [mx, mi] = sgt.query(l, r);
        cout << mx - mi << '\n';
    }
    return 0;
}

方法二

#include <bits/stdc++.h>
using namespace std;
using ll = long long;

struct T {
    int mx, mi;
    static T e() { return { (int)-2e9,(int)2e9 }; }
    friend T operator+(const T &a, const T &b) { return { max(a.mx, b.mx),min(a.mi,b.mi) }; }
};

template<class T>
class ST {
    vector<vector<T>> node;

public:
    ST() {}
    ST(const vector<T> &src) { init(src); }

    void init(const vector<T> &src) {
        assert(src.size());
        int n = src.size() - 1;
        int sz = log2(n);
        node.assign(sz + 1, vector<T>(n + 1));
        for (int i = 1;i <= n;i++) node[0][i] = src[i];
        for (int i = 1;i <= sz;i++)
            for (int j = 1;j + (1 << i) - 1 <= n;j++)
                node[i][j] = node[i - 1][j] + node[i - 1][j + (1 << i - 1)];
    }

    T query(int l, int r) {
        int k = log2(r - l + 1);
        return node[k][l] + node[k][r - (1 << k) + 1];
    }
};

int main() {
    std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int n, q;
    cin >> n >> q;
    vector<T> a(n + 1);
    for (int i = 1, x;i <= n;i++) cin >> x, a[i] = { x,x };
    ST<T> st(a);
    while (q--) {
        int l, r;
        cin >> l >> r;
        auto [mx, mi] = st.query(l, r);
        cout << mx - mi << '\n';
    }
    return 0;
}