[线段树][简单复杂度分析]LOJ#6029. 「雅礼集训 2017 Day1」市场

最新推荐文章于 2022-03-18 13:12:03 发布

Vectorxj

最新推荐文章于 2022-03-18 13:12:03 发布

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分类专栏：复杂度分析线段树

本文链接：https://blog.youkuaiyun.com/Vectorxj/article/details/78231804

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复杂度分析

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本文介绍了一种基于线段树的区间操作与查询算法，包括区间整除、区间加法、区间最小值查询及区间求和等核心功能，并提供了一个完整的C++实现示例。

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$Description$

要求支持区间整除一个正整数，区间加，区间询问最小值，区间求和。

$Solution$

就是复杂度分析吧。。
和以前一道区间开根号的题差不多吧。。
好菜啊。只会做水题了。。

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

const int N = 101010;
const long long INF = 1ll << 60;
typedef long long ll;

inline char get(void) {
    static char buf[100000], *S = buf, *T = buf;
    if (S == T) {
        T = (S = buf) + fread(buf, 1, 100000, stdin);
        if (S == T) return EOF;
    }
    return *S++;
}
template<typename T>
inline void read(T &x) {
    static char c; x = 0; int sgn = 0;
    for (c = get(); c < '0' || c > '9'; c = get()) if (c == '-') sgn = 1;
    for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';
    if (sgn) x = -x;
}

int a[N];
ll sum[N << 2], mx[N << 2], mn[N << 2], add[N << 2];
int n, q, opt, l, r, x;

inline void PushDown(int o, int l, int r) {
    if (add[o]) {
        int mid = (l + r) >> 1;
        add[o << 1] += add[o]; add[o << 1 | 1] += add[o];
        mx[o << 1] += add[o]; mx[o << 1 | 1] += add[o];
        mn[o << 1] += add[o]; mn[o << 1 | 1] += add[o];
        sum[o << 1] += add[o] * (mid + 1 - l);
        sum[o << 1 | 1] += add[o] * (r - mid);
        add[o] = 0;
    }
}
inline void PushUp(int o) {
    mx[o] = max(mx[o << 1], mx[o << 1 | 1]);
    mn[o] = min(mn[o << 1], mn[o << 1 | 1]);
    sum[o] = sum[o << 1] + sum[o << 1 | 1];
}
inline ll Div(ll x, ll y) {
    return floor((double)x / y);
}
inline void Build(int o, int l, int r) {
    add[o] = 0;
    if (l == r) {
        sum[o] = mx[o] = mn[o] = a[l];
        return;
    }
    int mid = (l + r) >> 1;
    Build(o << 1, l, mid);
    Build(o << 1 | 1, mid + 1, r);
    PushUp(o);
}
inline void Add(int o, int l, int r, int L, int R, int x) {
    if (l >= L && r <= R) {
        add[o] += x; mx[o] += x; mn[o] += x;
        sum[o] += x * (r - l + 1);
        return;
    }
    PushDown(o, l, r);
    int mid = (l + r) >> 1;
    if (L <= mid) Add(o << 1, l, mid, L, R, x);
    if (R > mid) Add(o << 1 | 1, mid + 1, r, L, R, x);
    PushUp(o);
}
inline void Div(int o, int l, int r, int L, int R, int d) {
    if (l >= L && r <= R && mx[o] - Div(mx[o], d) == mn[o] - Div(mn[o], d)) {
        ll D = Div(mx[o], d) - mx[o];
        add[o] += D; mx[o] += D; mn[o] += D;
        sum[o] += D * (r - l + 1);
        return;
    }
    PushDown(o, l, r);
    int mid = (l + r) >> 1;
    if (L <= mid) Div(o << 1, l, mid, L, R, d);
    if (R > mid) Div(o << 1 | 1, mid + 1, r, L, R, d);
    PushUp(o);
}
inline ll Sum(int o, int l, int r, int L, int R) {
    if (l >= L && r <= R) return sum[o];
    PushDown(o, l, r);
    int mid = (l + r) >> 1; ll res = 0;
    if (L <= mid) res += Sum(o << 1, l, mid, L, R);
    if (R > mid) res += Sum(o << 1 | 1, mid + 1, r, L, R);
    return res;
}
inline ll Min(int o, int l, int r, int L, int R) {
    if (l >= L && r <= R) return mn[o];
    PushDown(o, l, r);
    int mid = (l + r) >> 1; ll res = INF;
    if (L <= mid) res = min(res, Min(o << 1, l, mid, L, R));
    if (R > mid) res = min(res, Min(o << 1 | 1, mid + 1, r, L, R));
    return res;
}

int main(void) {
    freopen("1.in", "r", stdin);
    read(n); read(q);
    for (int i = 1; i <= n; i++) read(a[i]);
    Build(1, 1, n);
    for (int i = 1; i <= q; i++) {
        read(opt); read(l); read(r); ++l; ++r;
        if (opt == 1) {
            read(x); Add(1, 1, n, l, r, x);
        } else if (opt == 2) {
            read(x); Div(1, 1, n, l, r, x);
        } else if (opt == 3) {
            printf("%lld\n", Min(1, 1, n, l, r));
        } else {
            printf("%lld\n", Sum(1, 1, n, l, r));
        }
    }
    return 0;
}