Intel Code Challenge Elimination Round (Div.1 + Div.2, combined) [D] Destroying Array

Destroying Array

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

You are given an array consisting of n non-negative integers
a1, a2, ..., an.

You are going to destroy integers in the array one by one. Thus, you are given the permutation of integers from
1 to n defining the order elements of the array are destroyed.

After each element is destroyed you have to find out the segment of the array, such that it contains no destroyed elements and the sum of its elements is maximum possible. The sum of elements in the empty segment is considered to be
0.

Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the length of the array.

The second line contains n integers
a1, a2, ..., an (0 ≤ ai ≤ 109).

The third line contains a permutation of integers from
1 to n — the order used to destroy elements.

Output
Print n lines. The
i-th line should contain a single integer — the maximum possible sum of elements on the segment containing no destroyed elements, after first
i operations are performed.

Examples

Input

4
1 3 2 5
3 4 1 2

Output

5
4
3
0

Input

5
1 2 3 4 5
4 2 3 5 1

Output

6
5
5
1
0

Input

8
5 5 4 4 6 6 5 5
5 2 8 7 1 3 4 6

Output

18
16
11
8
8
6
6
0

Note
Consider the first sample:

Third element is destroyed. Array is now 1 3  *  5. Segment with maximum sum
5 consists of one integer 5.

Fourth element is destroyed. Array is now 1 3  *   * . Segment with maximum sum
4 consists of two integers 1 3.

First element is destroyed. Array is now  *  3  *   * . Segment with maximum sum
3 consists of one integer 3.

Last element is destroyed. At this moment there are no valid nonempty segments left in this array, so the answer is equal to
0.
这道题最开始是想的线段树，原因很简单:涉及到区间修改和区间询问

但是重测超时……，发现时间复杂度太高，常数太大，线段树肯定行不通。这题正解是并查集

首先将删除反着看成添加元素，对于每一次添加求出这个时候整个数组的最大连续和

因此将每个数组的位置看做一个点，讨论每一个点的时候就涉及到其相邻两个数，对于第i个点，如果i+1号点被添加到集合中，就将i号点的父结点改为i+1；如果i-1号点被添加到集合中，就将i号点的父结点改为i-1

#include<cstdio>
#include<iostream>
#define LL long long
using namespace std;
const LL maxn=100005;
inline void _read(LL &x){
char t=getchar();bool sign=true;
while(t<'0'||t>'9')
{if(t=='-')sign=false;t=getchar();}
for(x=0;t>='0'&&t<='9';t=getchar())x=x*10+t-'0';
if(!sign)x=-x;
}
LL n,s[maxn],v[maxn],fa[maxn],q[maxn],intree[maxn],ans[maxn];
LL getfa(LL x){return x==fa[x]?x:fa[x]=getfa(fa[x]);}
void merge(LL x,LL y){
LL fx=getfa(x),fy=getfa(y);
fa[fx]=fy;
v[fy]+=v[fx];
}
int main(){
_read(n);
LL i,j;
for(i=1;i<=n;i++)_read(s[i]);
for(i=1;i<=n;i++)_read(q[i]);
for(i=1;i<=n;i++)fa[i]=i;
for(i=n;i>1;i--){
LL cur=q[i];
v[cur]=s[cur];
if(intree[cur-1])merge(cur,cur-1);
if(intree[cur+1])merge(cur,cur+1);
ans[i-1]=max(ans[i],v[getfa(cur)]);
intree[cur]=1;
}
for(i=1;i<=n;i++)printf("%I64d\n",ans[i]);
}