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;
}