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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 andlf?=?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 integerm (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 meansrmq 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 usecout (also you may use %I64d).
Sample Input
41 2 3 443 03 0 -10 12 1
100
题意很好理解。
如果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,判断m个询问里是否存在c,详见代码,很好理解。
#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; }