干草堆计数(Counting Haybale P) 题解

思路分析

按题目要求使用对应数据结构与算法模拟/递推求解，注意数据范围边界与时间复杂度。

参考代码

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

const long long INF = (1LL << 60);
int n, q;
vector<long long> a, sumT, minT, tag;

void pull(int u) {
    sumT[u] = sumT[u << 1] + sumT[u << 1 | 1];
    minT[u] = min(minT[u << 1], minT[u << 1 | 1]);
}

void apply(int u, int l, int r, long long v) {
    sumT[u] += v * (r - l + 1);
    minT[u] += v;
    tag[u] += v;
}

void push(int u, int l, int r) {
    if (tag[u] != 0) {
        int mid = (l + r) >> 1;
        apply(u << 1, l, mid, tag[u]);
        apply(u << 1 | 1, mid + 1, r, tag[u]);
        tag[u] = 0;
    }
}

void build(int u, int l, int r) {
    if (l == r) {
        sumT[u] = minT[u] = a[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(u << 1, l, mid);
    build(u << 1 | 1, mid + 1, r);
    pull(u);
}

void update(int u, int l, int r, int ql, int qr, long long v) {
    if (ql <= l && r <= qr) {
        apply(u, l, r, v);
        return;
    }
    push(u, l, r);
    int mid = (l + r) >> 1;
    if (ql <= mid) update(u << 1, l, mid, ql, qr, v);
    if (qr > mid) update(u << 1 | 1, mid + 1, r, ql, qr, v);
    pull(u);
}

long long queryMin(int u, int l, int r, int ql, int qr) {
    if (ql <= l && r <= qr) return minT[u];
    push(u, l, r);
    int mid = (l + r) >> 1;
    long long res = INF;
    if (ql <= mid) res = min(res, queryMin(u << 1, l, mid, ql, qr));
    if (qr > mid) res = min(res, queryMin(u << 1 | 1, mid + 1, r, ql, qr));
    return res;
}

long long querySum(int u, int l, int r, int ql, int qr) {
    if (ql <= l && r <= qr) return sumT[u];
    push(u, l, r);
    int mid = (l + r) >> 1;
    long long res = 0;
    if (ql <= mid) res += querySum(u << 1, l, mid, ql, qr);
    if (qr > mid) res += querySum(u << 1 | 1, mid + 1, r, ql, qr);
    return res;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    if (!(cin >> n >> q)) return 0;
    a = vector<long long>(n + 1, 0);
    for (int i = 1; i <= n; ++i) cin >> a[i];
    sumT.assign(4 * n + 5, 0);
    minT.assign(4 * n + 5, INF);
    tag.assign(4 * n + 5, 0);
    build(1, 1, n);
    for (int i = 0; i < q; ++i) {
        char op;
        int l, r;
        cin >> op >> l >> r;
        if (op == 'P') {
            long long c;
            cin >> c;
            update(1, 1, n, l, r, c);
        } else if (op == 'M') {
            cout << queryMin(1, 1, n, l, r) << "\n";
        } else {
            cout << querySum(1, 1, n, l, r) << "\n";
        }
    }
    return 0;
}