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:

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.

题意：有n个数，然后每个数的权值是a[i]，现在按照顺序去摧毁n个元素，然后每次问你最大的连通块和是多少。
解法：倒着做，然后用带权并查集去维护就好了。

//CF 722 C
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5+7;
typedef long long LL;
LL a[maxn], b[maxn], vis[maxn], sum[maxn], ans[maxn];
int n;
namespace dsu{
    int fa[maxn];
    inline int find_set(int x){if(x == fa[x]) return x; return fa[x] = find_set(fa[x]);}
    inline void union_set(int x, int y){int fx = find_set(x), fy = find_set(y); fa[fx] = fy; sum[fy] += sum[fx];}
}
using namespace dsu;
int main()
{
    scanf("%d", &n);
    for(int i = 1; i <= n; i++){
        scanf("%d", &a[i]);
        sum[i] = a[i];
        fa[i] = i;
    }
    for(int i = 1; i <= n; i++) scanf("%lld", &b[i]);
    LL now = 0;
    for(int i = n; i > 1; i--){
        vis[b[i]] = 1;
        if(vis[b[i]-1]) union_set(b[i]-1, b[i]);
        if(vis[b[i]+1]) union_set(b[i]+1, b[i]);
        now = max(now, sum[find_set(b[i])]);
        ans[i - 1] = now;
    }
    for(int i = 1; i <= n; i++) printf("%lld\n", ans[i]);
    return 0;
}