Codeforces 52C (线段树区间更新)

C. Circular RMQ

time limit per test
3 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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), aiare
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 test(s)

input

4
1 2 3 4
4
3 0
3 0 -1
0 1
2 1

output

1
0
0

题目链接：http://codeforces.com/problemset/problem/52/C

题目大意：求环型数组的rmq

题目分析：裸的区间更新

#include <cstdio>
#include <cstring>
#include <algorithm>
#define ll long long
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
using namespace std;
int const MAX = 200005;
int const INF = 0x3fffffff;
char s[100];
int mi[MAX << 2], lazy[MAX << 2];
int n, q;

void PushUp(int rt)
{
mi[rt] = min(mi[rt << 1], mi[rt << 1 | 1]);
}

void PushDown(int rt)
{
if(lazy[rt])
{
mi[rt << 1] += lazy[rt];
mi[rt << 1 | 1] += lazy[rt];
lazy[rt << 1] += lazy[rt];
lazy[rt << 1 | 1] += lazy[rt];
lazy[rt] = 0;
}
}

void Build(int l, int r, int rt)
{
lazy[rt] = 0;
if(l == r)
{
scanf("%d", &mi[rt]);
return;
}
int mid = (l + r) >> 1;
Build(lson);
Build(rson);
PushUp(rt);
}

void Update(int L, int R, int c, int l, int r, int rt)
{
if(L <= l && r <= R)
{
mi[rt] += c;
lazy[rt] += c;
return;
}
int mid = (l + r) >> 1;
PushDown(rt);
if(L <= mid)
Update(L, R, c, lson);
if(mid < R)
Update(L, R, c, rson);
PushUp(rt);
}

int Query(int L, int R, int l, int r, int rt)
{
if(L <= l && r <= R)
return mi[rt];
int mid = (l + r) >> 1;
int ans = INF;
PushDown(rt);
if(L <= mid)
ans = min(ans, Query(L, R, lson));
if(mid < R)
ans = min(ans, Query(L, R, rson));
return ans;
}

int main()
{
scanf("%d", &n);
Build(1, n, 1);
scanf("%d", &q);
while(q --)
{
int l, r, c;
scanf("%d %d", &l, &r);
l ++;
r ++;
if(getchar() == ' ')
{
scanf("%d", &c);
if(l <= r)
Update(l, r, c, 1, n, 1);
else
{
Update(1, r, c, 1, n, 1);
Update(l, n, c, 1, n, 1);
}
}
else
{
if(l <= r)
printf("%d\n", Query(l, r, 1, n, 1));
else
printf("%d\n", min(Query(1, r, 1, n, 1), Query(l, n, 1, n, 1)));
}
}
}