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

C. 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:

1. Third element is destroyed. Array is now 1 3  *  5. Segment with maximum sum 5 consists of one integer 5.
2. Fourth element is destroyed. Array is now 1 3  *   * . Segment with maximum sum 4 consists of two integers 1 3.
3. First element is destroyed. Array is now  *  3  *   * . Segment with maximum sum 3 consists of one integer 3.
4. Last element is destroyed. At this moment there are no valid nonempty segments left in this array, so the answer is equal to 0.

题意：n个数，删除点的顺序，求每次最大的连续和；

思路：并查集，倒着写；线段树也可以；

#include
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
const int N=2e5+10,M=4e6+10,inf=1e9+10,mod=1e9+7;
const ll INF=1e18+10;
int fa[N],flag[N];
ll sum[N],ans[N],a[N],b[N];
int fa18(int x)
{
    return x==fa[x]?x:fa[x]=fa18(fa[x]);
}
void update(int u,int v,int i)
{
    int x=fa18(u);
    int y=fa18(v);
    if(x!=y)
    {
        fa[x]=y;
        sum[y]+=sum[x];
        ans[i]=max(ans[i+1],max(ans[i],sum[y]));
    }
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    for(int i=1;i<=n;i++)
        scanf("%d",&b[i]);
    for(int i=1;i<=n;i++)
        fa[i]=i,sum[i]=a[i];
    for(int i=n;i>=1;i--)
    {
        if(b[i]==fa[b[i]])
        ans[i]=max(ans[i+1],max(sum[b[i]],ans[i]));
        if(flag[b[i]-1])update(b[i]-1,b[i],i);
        if(flag[b[i]+1])update(b[i]+1,b[i],i);
        flag[b[i]]=1;
    }
    for(int i=2;i<=n+1;i++)
    {
        printf("%lld ",ans[i]);
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/jhz033/p/5927837.html