Codeforces 272C Dima and Staircase 线段树区间覆盖，最值查询

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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 heighta₁, the second one is ata₂, the last one is ata_n (1 ≤ a₁ ≤ a₂ ≤ ... ≤ a_n).

Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. Thei-th box has width w_i and heighth_i. Dima throws each box vertically down on the firstw_i stairs of the staircase, that is, the box covers stairs with numbers1, 2, ..., w_i. 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 widthw_i cannot touch the stair numberw_i + 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 ofn integers, a₁, a₂, ..., a_n(1 ≤ a_i ≤ 10⁹; a_i ≤ a_i + 1).

The next line contains integer m (1 ≤ m ≤ 10⁵) — the number of boxes. Each of the followingm lines contains a pair of integers w_i, h_i(1 ≤ w_i ≤ n; 1 ≤ h_i ≤ 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 inC++. It is preferred to use the cin, cout streams or the %I64d specifier.

Sample test(s)

Input

51 2 3 6 641 13 11 14 3

Output

1346

Input

31 2 321 13 1

Output

13

Input

1151 21 101 101 101 10

Output

13132333

Note

The first sample are shown on the picture.

题意：从左往右，给出每个位置的楼梯高度，依次垂直第在最左边放上高度为hi,宽度为wi的箱子，如上图，箱子不能进行选择，对于每一个箱子，输出这个箱子所在的高度。

思路：对于每一个箱子，只需要查询[1,Wi]内的最大值，箱子会被这个最大值支撑着，所以输出这个最大值，放上这个箱子之后，这个区间内的所有值都会变成上述的那个最大值加上箱子的高度Hi。用线段树维护即可。

#include<cstdio>#include<cstring>#include<algorithm>#define LL long long#define maxn 100005using namespace std;struct node{int l, r;LL c, maxh;}tree[maxn << 2];int h[maxn];void pushdown(int id){if (tree[id].c){tree[id << 1].c = tree[id].c;tree[id << 1 | 1].c = tree[id].c;tree[id << 1].maxh = tree[id].maxh;tree[id << 1 | 1].maxh = tree[id].maxh;}}void build(int id, int l, int r){tree[id].l = l;tree[id].r = r;if (tree[id].l == tree[id].r){tree[id].maxh = h[l];}else{int mid = (tree[id].l + tree[id].r) >> 1;build(id << 1, l, mid);build(id << 1 | 1, mid + 1, r);tree[id].maxh = max(tree[id << 1].maxh, tree[id << 1 | 1].maxh);}}void op(int id, int l, int r, LL c){if (l <= tree[id].l&&tree[id].r <= r){tree[id].c = c;tree[id].maxh = c;}else{pushdown(id);int mid = (tree[id].l + tree[id].r) >> 1;if (l <= mid) op(id << 1, l, r, c);if (mid<r) op(id << 1 | 1, l, r, c);tree[id].maxh = max(tree[id << 1].maxh, tree[id << 1 | 1].maxh);}}LL que(int id, int l, int r){if (l <= tree[id].l&&tree[id].r <= r){return tree[id].maxh;}else{pushdown(id);int mid = (tree[id].l + tree[id].r) >> 1;LL res = 0;if (l <= mid) res = max(res, que(id << 1, l, r));if (mid<r) res = max(res, que(id << 1 | 1, l, r));return res;}}int main(){int n;scanf("%d", &n);for (int i = 1; i <= n; i++) scanf("%d", &h[i]);int m;scanf("%d", &m);build(1, 1, n);while (m--){LL w, h;scanf("%I64d %I64d", &w, &h);LL tmp = que(1, 1, w);printf("%I64d\n", tmp);op(1, 1, w, tmp + h);}return 0;}