Codeforces 52C

C. Circular RMQ

题目大意：

好激动，带懒标记的上百行的线段树带A掉这道题,线段树真好玩，可以看出来是裸的带懒标记的线段树
难点1：读入的时候怎么判断是3个数还是2个数，看快读最后有无读入空格，如果读入了space就等于1
难点2：线段是环形的l <= r正常区间，l > r 就讲区间分成 1 ~ r, l ~ n, 读入的是0~n-1我们处理成1 ~ n
难点3：线段树和懒标记，我也是初学者不会讲，找大佬学呗

Ac code

#include <bits/stdc++.h>
using namespace std;typedef long long ll;
const int N = 2e5 + 10;
ll w[N];
struct Node{ll l, r, minv, lazy;
}tr[N << 2];int space;
inline ll read()
{space = 0;ll x = 0, f = 1;char c = getchar();while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9'){x = x * 10 + c - '0';c = getchar();}space = c == ' ';return x * f;
}inline void print(ll x){if(x < 0){putchar('-');x = -x;}if(x > 9) print(x / 10);putchar(x % 10 + '0');
}inline ll ls(ll u) {return u << 1; }
inline ll rs(ll u) {return u << 1 | 1; }inline void pushup(ll u, ll l, ll r){tr[u] = {l, r, min(tr[ls(u)].minv, tr[rs(u)].minv), 0};
}inline void pushdown(ll u){if(tr[u].lazy){ //懒标记tr[ls(u)].minv += tr[u].lazy;tr[rs(u)].minv += tr[u].lazy;tr[ls(u)].lazy += tr[u].lazy;tr[rs(u)].lazy += tr[u].lazy;tr[u].lazy = 0;}
}void build(ll u, ll l, ll r)
{if(l == r){tr[u] = {l, r, w[l], 0};return;}ll mid = l + r >> 1;build(ls(u), l, mid);build(rs(u), mid + 1, r);pushup(u, l, r);
}void modify(ll u, ll l, ll r, ll x)
{if(l <= tr[u].l && tr[u].r <= r){tr[u].minv += x;tr[u].lazy += x;return;}pushdown(u);ll mid = tr[u].l + tr[u].r >> 1;if(l <= mid) modify(ls(u), l, r, x);if(mid < r)  modify(rs(u), l, r, x);pushup(u, tr[u].l, tr[u].r);
}ll query(ll u, ll l, ll r)
{if(l <= tr[u].l && tr[u].r <= r)return tr[u].minv;pushdown(u);ll mid = tr[u].l + tr[u].r >> 1, minv = __INT64_MAX__;if(l <= mid) minv = min(minv, query(ls(u), l, r));if(mid < r)  minv = min(minv, query(rs(u), l, r));return minv;
}int main()
{#ifdef LOCALfreopen("in.txt", "r", stdin);#endifll n = read();for(int i = 1; i <= n; ++i) w[i] = read();build(1, 1, n); //建树ll m = read();    while(m--){ll x = read(), y = read();if(space) //inc{ll v = read();++x, ++y;if(x <= y)modify(1, x, y, v);else modify(1, 1, y, v), modify(1, x, n, v);}else    //rmq{++x, ++y;if(x <= y){print(query(1, x, y));putchar('\n');}else{print(min(query(1, 1, y), query(1, x, n)));putchar('\n');}}}return 0;
}

今天是被大运河虐惨的一天… ---------------------------------------------------------------------------------------------2021.12.18

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Codeforces 52C

C. Circular RMQ

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