Codeforces 52C (線段樹區間更新)

_TCgogogo_發表於2015-11-28
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
題目連結：http://codeforces.com/problemset/problem/52/C

題目大意：求環型陣列的rmq

題目分析：裸的區間更新

#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)));
		}
	}
}




相關文章