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CodeForces - 272C Dima and Staircase （线段树区间更新）

發布時間：2024/1/8 编程问答 28 豆豆

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題意：

? ? ? ? 見以下樣例，給出 5 個區間，每個區間的高度已知。一共 4 次操作。每次操作都是從最左邊開始向下壘一個寬為 w 高為h 的木塊，過程見下圖。

? ? ? ? 問每次壘木塊的高度是多少？

Input 5 1 2 3 6 6 4 1 1 3 1 1 1 4 3 Output 1 3 4 6 The first sample are shown on the picture.

思路：

? ? ? ? 線段樹區間更新裸題，注意爆int

代碼：

#include <bits/stdc++.h>using namespace std; #define mi (l+r)/2; #define ls l,mid,rt*2 #define rs mid+1,r,rt*2+1 const int MAXN=1e5; long long tree[MAXN*4],lazy[MAXN*4]; long long n,m,st,en,ans,w,h; void push_down(int rt){if(lazy[rt]){lazy[rt*2]=lazy[rt*2+1]=lazy[rt];tree[rt*2]=tree[rt*2+1]=lazy[rt];lazy[rt]=0;}return ; } void push_up(int rt){tree[rt]=max(tree[rt*2],tree[rt*2+1]);return ; } void build(int l,int r,int rt){if(l==r){scanf("%I64d",&tree[rt]);return ;}int mid=mi;build(ls);build(rs);push_up(rt);return ; } void update(int l,int r,int rt){if(st<=l&&r<=en){tree[rt]=lazy[rt]=ans;return ;}int mid=mi;push_down(rt);if(st<=mid) update(ls);if(mid<en) update(rs);push_up(rt); } void query(int l,int r,int rt){if(st<=l&&r<=en){ans=max(tree[rt],ans);return ;}int mid=mi;push_down(rt);if(st<=mid) query(ls);if(mid<en) query(rs);return ; } int main() {while(scanf("%I64d",&n)!=-1){memset(tree,0,sizeof(tree));memset(lazy,0,sizeof(lazy));build(1,n,1);scanf("%I64d",&m);st=1;while(m--){scanf("%I64d%I64d",&w,&h);en=w;ans=0;query(1,n,1);printf("%I64d\n",ans);ans+=h;update(1,n,1);}} }

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?a_n?(1?≤?a₁?≤?a₂?≤?...?≤?a_n).

Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The?i-th box has width?w_i?and height?h_i. Dima throws each box vertically down on the first?w_i?stairs of the staircase, that is, the box covers stairs with numbers?1,?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 width?w_i?cannot touch the stair number?w_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 of?nintegers,?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 following?m?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 in?C++. It is preferred to use the?cin,?cout?streams or the?%I64d?specifier.

Example

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.

總結

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