Codeforces 272C Dima and Staircase (線段樹區間更新 或 線性掃)

_TCgogogo_發表於2015-11-28
AI
C. Dima and Staircase
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an(1 ≤ a1 ≤ a2 ≤ ... ≤ an).

Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:

  • the bottom of the box touches the top of a stair;
  • the bottom of the box touches the top of a box, thrown earlier.

We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.

You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.

Input

The first line contains integer n (1 ≤ n ≤ 105) — the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 ≤ ai ≤ 109; ai ≤ ai + 1).

The next line contains integer m (1 ≤ m ≤ 105) — the number of boxes. Each of the following m lines contains a pair of integers wi, hi(1 ≤ wi ≤ n; 1 ≤ hi ≤ 109) — the size of the i-th thrown box.

The numbers in the lines are separated by spaces.

Output

Print m integers — for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64dspecifier.

Sample test(s)
input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
output
1
3
4
6
input
3
1 2 3
2
1 1
3 1
output
1
3
input
1
1
5
1 2
1 10
1 10
1 10
1 10
output
1
3
13
23
33
Note

The first sample are shown on the picture.

題目連結：http://codeforces.com/contest/272/problem/C

題目大意：直接看圖就知道了吧

題目分析：div2的C能有多難?，由於這題有個特殊性，要放必須從左端點開始放，因此可以線性維護左端點到放的位置的最高點，但是既然是在練習線段樹，我們就要用線段樹的思想來思考這個問題，實質上還是維護一個區間的最高點，如果說不一定要從左端點開始放，那就只能線段樹了
線段樹區間更新
#include <cstdio>
#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 = 1e5 + 5;
ll h[MAX << 2], lazy[MAX << 2];
int n, q;

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

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

void Build(int l, int r, int rt)
{
	lazy[rt] = 0;
	if(l == r)
	{
		scanf("%I64d", &h[rt]);
		return;
	}
	int mid = (l + r) >> 1;
	Build(lson);
	Build(rson);
	PushUp(rt);
}

void Update(int L, int R, ll c, int l, int r, int rt)
{
	if(L <= l && r <= R)
	{
		h[rt] = c;
		lazy[rt] = h[rt];
		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);
	return;
}

ll Query(int L, int R, int l, int r, int rt)
{
	if(L <= l && r <= R)
		return h[rt];
	int mid = (l + r) >> 1;
	PushDown(rt);
	ll ans = 0;
	if(L <= mid)
		ans = max(ans, Query(L, R, lson));
	if(mid < R)
		ans = max(ans, Query(L, R, rson));
	return ans;
}

int main()
{
	scanf("%d", &n);
	Build(1, n, 1);
	scanf("%d", &q);
	while(q--)
	{
		int wd, ht;
		scanf("%d %d", &wd, &ht);
		int ll ans = Query(1, wd, 1, n, 1);
		printf("%I64d\n", ans);
		Update(1, wd, ans + (ll)ht, 1, n, 1);
	}
}

線性掃
#include <cstdio>
#include <algorithm>
#define ll long long
using namespace std;
int const MAX = 1e5 + 5;
ll h[MAX];

int main()
{	
	int n, q;
	scanf("%d", &n);
	for(int i = 1; i <= n; i++)
		scanf("%I64d", &h[i]);
	ll curh = 0;
	scanf("%d", &q);
	while(q --)
	{
		int w, hh;
		scanf("%d %d", &w, &hh);
		ll ma = max(curh, h[w]);
		printf("%I64d\n", ma);
		curh = ma + (ll)hh;
	}
}



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