[CF52C]Circular RMQ【线段树】

题目大意

给你一个环形数列，完成环形数列上区间加法和区间求最小值。

分析

算是一道比较水的线段树模板题。
如果l>r的话，那么修改l,n和1,r区间。
不然的话那么就修改l,r区间。
其他的基础操作可以看我的随笔：【传送门】
还要注意开long long。

ac代码

#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define N 200005
#define ll long long
#define linf 9223372036854775807
using namespace std;
ll a[N];
bool bo;
template <typename T>
inline void read(T &x) {
    x = 0; T fl = 1;
    bo = 0;
    char ch = 0;
    while (ch < '0' || ch > '9') {
        if (ch == '-') fl = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
        x = (x << 1) + (x << 3) + (ch ^ 48);
        ch = getchar();
    }
    if (ch == ' ') bo = 1;
    x *= fl;
}
struct segment_tree {
    #define lc (nod << 1)
    #define rc (nod << 1 | 1)
    #define mid ((l + r) >> 1)
    struct node {
        ll mn, tag;
        int l, r;
        node() {
            mn = l = r = tag;
        }
    }tr[N << 2];
    void pushup(int nod) {
        tr[nod].mn = min(tr[lc].mn, tr[rc].mn);
    }
    void pushdown(int nod) {
        if (tr[nod].tag == 0 || tr[nod].l == tr[nod].r) return;
        ll tmp = tr[nod].tag;
        tr[nod].tag = 0;
        tr[lc].mn += tmp;
        tr[rc].mn += tmp;
        tr[lc].tag += tmp;
        tr[rc].tag += tmp;
    }
    void build(int l, int r, int nod, ll *a) {
        tr[nod].l = l, tr[nod].r = r;
        if (l == r) {
            tr[nod].mn = a[l];
            return;
        }
        build(l, mid, lc, a);
        build(mid + 1, r, rc, a);
        pushup(nod);
    }
    void update_sec(int ql, int qr, int nod, ll val) {
        int l = tr[nod].l, r = tr[nod].r;
        pushdown(nod);
        if (ql <= l && r <= qr) {
            tr[nod].mn += val;
            tr[nod].tag += val;
            return;
        }
        if (ql <= mid) update_sec(ql, qr, lc, val);
        if (qr > mid) update_sec(ql, qr, rc, val);
        pushup(nod);
    }
    ll query_sec_min(int ql, int qr, int nod) {
        int l = tr[nod].l, r = tr[nod].r;
        pushdown(nod);
        if (ql <= l && r <= qr) return tr[nod].mn;
        ll res = linf;
        if (ql <= mid) res = min(res, query_sec_min(ql, qr, lc));
        if (qr > mid) res = min(res, query_sec_min(ql, qr, rc));
        return res;
    }
}tr;
int n, m;
int main() {
    read(n);
    for (int i = 1; i <= n; i ++) read(a[i]);
    tr.build(1, n, 1, a);
    read(m);
    while (m --) {
        int l, r;
        read(l); read(r);
        l ++; r ++;
        if (!bo) {
            ll res = linf;
            if (l > r) {
                res = min(res, tr.query_sec_min(l, n, 1));
                res = min(res, tr.query_sec_min(1, r, 1));
            }
            else res = min(res, tr.query_sec_min(l, r, 1));
            printf("%lld\n", res);
        }
        else {
            ll x;
            cin>> x;
            if (l > r) {
                tr.update_sec(l, n, 1, x);
                tr.update_sec(1, r, 1, x);
            }
            else tr.update_sec(l, r, 1, x);
        }
    }
    return 0;
}