「CF52C」Circular RMQ

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Portal

Portal1: Codeforces

Portal2: Luogu

Description

You are given circular array $a_0, a_1, \cdots, a_{n - 1}$. There are two types of operations with it:

$\textrm{inc}(lf, rg, v)$ — this operation increases each element on the segment $[lf, rg]$ (inclusively) by $v$;
$\textrm{rmq}(lf, rg)$ — this operation returns minimal value on the segment $[lf, rg]$ (inclusively).

Assume segments to be circular, so if $n = 5$ and $lf = 3, rg = 1$, it means the index sequence: $3, 4, 0, 1$.

Write program to process given sequence of operations.

Input

The first line contains integer $n (1 \le n \le 200000)$. The next line contains initial state of the array: $a_0, a_1, \cdots, a_{n - 1} ( -10^6 \le ai \le 10^6)$, $a_i$ are integer. The third line contains integer $m (0 \le m \le 200000)$, $m$ — the number of operartons. Next $m$ lines contain one operation each. If line contains two integer $lf, rg (0 \le lf, rg \le n - 1)$ it means rmq operation, it contains three integers $lf, rg, v (0 \le lf, rg \le n - 1; -10^6 \le v \le 10^6)$ — inc operation.

Output

For each rmq operation write result for it. Please, do not use %lld specificator to read or write $64$-bit integers in C++. It is preffered to use cout (also you may use %I64d).

Sample Input

Sample Output

Solution

我们可以用线段树来解决区间RMQ问题，我们在线段树上维护一个最小值与懒标记，这样问题就解决了。

读入的时候我们可以判断后面一个字符是不是空格，可以直接在快速读入里判断，这样就可以判断出一行有三个数还是两个数。

Code

#include<iostream>

#include<algorithm>

#include<cstdio>

#include<cstring>

#include<cmath>

using namespace std;

const int MAXN = 200005;

int n, m, l, r, val, a[MAXN];

bool opt;

namespace Segtree {

    #define ls rt << 1

    #define rs rt << 1 | 1

    typedef long long LL;

    const LL Seg_INF = 1e18;

    const int Seg_MAXN = 1000005;

    struct SMT {

        LL Min, tag;

    } tree[Seg_MAXN];

    inline void build(int rt, int l, int r) {//建立线段树

        if (l == r) {

            tree[rt].Min = a[l];

            return ;

        }

        int mid = l + r >> 1;

        build(ls, l, mid);

        build(rs, mid + 1, r);

        tree[rt].Min = min(tree[ls].Min, tree[rs].Min);

    }

    inline void update(int rt, int l, int r, int ansl, int ansr, int val) {//线段树修改

        if (ansl <= l && r <= ansr) {

            tree[rt].tag += val;

            return ;

        }

        int mid = l + r >> 1;

        if (ansl <= mid) update(ls, l, mid, ansl, ansr, val);

        if (mid < ansr) update(rs, mid + 1, r, ansl, ansr, val);

        tree[rt].Min = min(tree[ls].Min + tree[ls].tag, tree[rs].Min + tree[rs].tag);

    }

    inline LL query(int rt, int l, int r, int ansl, int ansr) {//线段树查询

        if (ansl <= l && r <= ansr) return tree[rt].Min + tree[rt].tag;

        int mid = l + r >> 1;

        LL ret = Seg_INF;

        if (ansl <= mid) ret = min(ret, query(ls, l, mid, ansl, ansr));

        if (mid < ansr) ret = min(ret, query(rs, mid + 1, r, ansl, ansr));

        return ret + tree[rt].tag;

    }

}

using namespace Segtree;

inline int read() {

    opt = 0;

    char ch = getchar();

    int x = 0, f = 1;

    while (ch < '0' || ch > '9') {

        if (ch == '-') f = -1;

        ch = getchar();

    }

    while ('0' <= ch && ch <= '9') {

        x = (x << 1) + (x << 3) + ch - '0';

        ch = getchar();

    }

    if (ch == ' ') opt = 1;//判断空格

    return x * f;

}

int main() {

    n = read();

    for (int i = 1; i <= n; i++)

        a[i] = read();

    build(1, 1, n);

    m = read();

    for (int i = 1; i <= m; i++) {

        l = read(); r = read(); l++; r++;

        if (!opt) {

            if (l <= r) printf("%lld\n", query(1, 1, n, l, r)); else printf("%lld\n", min(query(1, 1, n, l, n), query(1, 1, n, 1, r)));

        } else {

            val = read();

            if (l <= r) update(1, 1, n, l, r, val); else {

                update(1, 1, n, l, n, val);

                update(1, 1, n, 1, r, val);

            }

        }

    }

    return 0;

}