【37.38%】【codeforces 722C】Destroying Array

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard 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.

【题解】



线段树区间合并

每次删掉一个节点;

类似hotel那题,不同的是变成求连续的块的最大和。

可以在求连续块的基础上做一些加法运算求和。

每次操作完更新完之后输出mlx[1]即可(根节点代表整个区间);

#include <cstdio>
#include <algorithm>

using namespace std;

const int MAXN = 109000;

int n, m;
int  llianxusum[MAXN<<2], rlianxusum[MAXN<<2];
long long mllx[MAXN<<2], mrlx[MAXN<<2], mlx[MAXN<<2];

void build(int l, int r, int rt)
{
    llianxusum[rt] = rlianxusum[rt] = r - l + 1;
    if (l == r)
    {
        scanf("%I64d", &mlx[rt]);
        mllx[rt] = mrlx[rt] = mlx[rt];
        return;
    }
    int mid = (l + r) >> 1;
    build(l, mid, rt << 1);
    build(mid + 1, r, rt << 1 | 1);
    mlx[rt] = mllx[rt] = mrlx[rt] = mlx[rt << 1] + mlx[rt << 1 | 1];
}

void push_up(int rt, int len)
{
    llianxusum[rt] = llianxusum[rt << 1];
    mllx[rt] = mllx[rt << 1];
    if (llianxusum[rt] == len - (len >> 1))
    {
        llianxusum[rt] += llianxusum[rt << 1 | 1];
        mllx[rt] += mllx[rt << 1 | 1];//连续的序列和也可以递增
    }
    rlianxusum[rt] = rlianxusum[rt << 1 | 1];
    mrlx[rt] = mrlx[rt << 1 | 1];
    if (rlianxusum[rt] == len >> 1)
    {
        rlianxusum[rt] += rlianxusum[rt << 1];
        mrlx[rt] += mrlx[rt << 1];//同理
    }
    mlx[rt] = max(mlx[rt << 1], mlx[rt << 1 | 1]);
    mlx[rt] = max(mlx[rt], mrlx[rt << 1] + mllx[rt << 1 | 1]);
    //整个区间的连续空格子数就没必要记录了。
    //直接整个区间的最大和就好。
}

void updata(int pos, int begin, int end, int rt)
{
    if (begin == end)
    {
        llianxusum[rt] = rlianxusum[rt] = 0;
        mlx[rt] = mllx[rt] = mrlx[rt] = 0;
        return;
    }
    int m = (begin + end) >> 1;
    if (pos <= m)
        updata(pos, begin, m, rt << 1);
    else
        updata(pos, m + 1, end, rt << 1 | 1);
    push_up(rt, end - begin + 1);
}

int main()
{
    //freopen("F:\\rush.txt", "r", stdin);
    //freopen("F:\\rush_out.txt", "w", stdout);
    scanf("%d", &n);
    build(1, n, 1);
    for (int i = 1; i <= n; i++)
    {
        int x;
        scanf("%d", &x);
        updata(x, 1, n, 1);
        printf("%I64d\n", mlx[1]);
    }
    return 0;
}