CF#52 C Circular RMQ (线段树区间更新)
Description
You are given circular array a0, a1, ..., an - 1. There are two types of operations with it:
- inc(lf, rg, v) — this operation increases each element on the segment [lf, rg] (inclusively) by v;
- 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 ≤ n ≤ 200000). The next line contains initial state of the array: a0, a1, ..., an - 1 ( - 106 ≤ ai ≤ 106), ai are integer. The third line contains integer m (0 ≤ m ≤ 200000), m — the number of operartons. Next m lines contain one operation each. If line contains two integer lf, rg (0 ≤ lf, rg ≤ n - 1) it means rmq operation, it contains three integers lf, rg, v (0 ≤ lf, rg ≤ n - 1; - 106 ≤ v ≤ 106) — 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
4 1 2 3 4 4 3 0 3 0 -1 0 1 2 1
1 0 0
题意非常好理解。
假设a>b的话,就查0~b和a~n-1,其余的就是线段树区间更新模板,推断m个询问里是否存在c,详见代码,非常好理解。
#include <stdio.h> #include <string.h> #include <algorithm> #include <math.h> #include <ctype.h> #include <iostream> #define lson o << 1, l, m #define rson o << 1|1, m+1, r using namespace std; typedef __int64 LL; const __int64 MAX = 9223372036854775807; const int maxn = 200010; int n, a, q, c, b; char str[1200]; LL mi[maxn<<2], add[maxn<<2]; void up(int o) { mi[o] = min(mi[o<<1], mi[o<<1|1]); } void down(int o) { if(add[o]) { add[o<<1] += add[o]; add[o<<1|1] += add[o]; mi[o<<1] += add[o]; mi[o<<1|1] += add[o]; add[o] = 0; } } void build(int o, int l, int r) { if(l == r) { scanf("%I64d", &mi[o]); return; } int m = (l+r) >> 1; build(lson); build(rson); up(o); } void update(int o, int l, int r) { if(a <= l && r <= b) { add[o] += c; mi[o] += c; return ; } down(o); int m = (l+r) >> 1; if(a <= m) update(lson); if(m < b ) update(rson); up(o); } LL query(int o, int l, int r) { if(a <= l && r <= b) return mi[o]; down(o); int m = (l+r) >> 1; LL res = MAX; if(a <= m) res = query(lson); if(m < b ) res = min(res, query(rson)); return res; } int main() { scanf("%d", &n); build(1, 0, n-1); scanf("%d", &q); getchar(); while(q--) { gets(str); if(sscanf(str,"%d %d %d", &a, &b, &c) == 3) { if(a <= b) update(1, 0, n-1); else { int tmp = b; b = n-1; update(1, 0, n-1); a = 0, b = tmp; update(1, 0, n-1); } } else { LL ans ; if(a <= b) ans = query(1, 0, n-1); else { int tmp = b; b = n-1; ans = query(1, 0, n-1); a = 0, b = tmp; ans = min(ans, query(1, 0, n-1)); } printf("%I64d\n", ans); } } return 0; }