codeforces(722c)-Destroying Array

原题链接

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.

定义两个集合set s1, multise s2, s1存的是破化的元素的下标，s2存的是区间的和。num[]数组存的是元素前缀和.每次破化一个元素k时，在s1中找到大于k最小的已被破化的元素下标p2, 和小于k最大的已被破化的下标p1, 那么k > p1 && k < p2 所以(p1, p2)区间被分成连个区间，把这两个区间的值存到s2中，删除(p1, p2)整个区间的值，输出s2中最大值

#include 
#include 
#include 
#include 
#include 
#include 
#define maxn 100005
using namespace std;
typedef long long ll;

ll num[maxn];
set s1;
multiset s2;
int main(){
	
//	freopen("in.txt", "r", stdin);
	int n, a;
	
	scanf("%d", &n);
	for(int i = 1; i <= n; i++){
		scanf("%I64d", num+i);
		num[i] += num[i-1];
	}
	set ::iterator iter;
	for(int i = 1; i <= n; i++){
		scanf("%d", &a);
		if(i == 1){
		   s1.insert(a);
		   s2.insert(num[n] - num[a]);
		   s2.insert(num[a-1]);
		   printf("%I64d\n", *(--s2.end()));
		}
		else{
			int h1, h2;
			iter = s1.upper_bound(a);
			if(iter == s1.end()){
				h2 = n;
				h1 = *(--iter);
			}
			else{
				h2 = *iter - 1;
				if(iter == s1.begin())
				 h1 = 0;
				else
				 h1 = *(--iter);
			}
			s1.insert(a);
			s2.insert(num[h2] - num[a]);
			s2.insert(num[a-1] - num[h1]);
			s2.erase(s2.find(num[h2] - num[h1]));
			printf("%I64d\n", *(--s2.end()));
		}	
	}
	return 0;
}