Destroying Array(并查集 高能坑点)

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

41 3 2 53 4 1 2

output

5430

input

51 2 3 4 54 2 3 5 1

output

65510

input

85 5 4 4 6 6 5 55 2 8 7 1 3 4 6

output

18161188660

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.

这道题可直接用并查集写，做法也比较多，当时我写出来查了一个小时愣是没找到错，有过了两天才明白错误在哪里，有可能很多点都为0，这样我的找祖先的策略就是错误的，还是用vis数组比较稳。

倒叙恢复每个点，并用并查集记录区间，记录恢复每个点后的最大值。看网上很多代码都是把当前点作为存储最大值的点，我是将最右端点作为存储最大值的点。附两个代码

AC代码①：

#include<cstdio>#include<iostream>#include<algorithm>#include<queue>#include<stack>#include<cstring>#include<string>#include<vector>#include<cmath>#include<map>using namespace std;typedef long long ll;#define mem(a,b) memset(a,b,sizeof(a))const int maxn = 2e5+5;const int ff = 0x3f3f3f3f;ll n,a[maxn],b[maxn];ll dp[maxn];ll ans[maxn];int ne[maxn];int f5(int x){return ne[x] == x?x:ne[x] = f5(ne[x]);//当时是这里写错了}int main(){cin>>n;mem(dp,0);for(int i = 0;i<= n+1;i++)ne[i] = i;for(int i = 1;i<= n;i++)scanf("%lld",&a[i]);for(int i = n;i>= 1;i--)scanf("%lld",&b[i]);ll maxx = -1;ans[1] = 0;for(int i = 1;i<= n-1;i++){dp[b[i]] = a[b[i]]+dp[b[i]-1];ne[b[i]-1] = b[i];int zy = f5(b[i]);if(zy!= b[i])dp[zy]+= dp[b[i]];maxx = max(maxx,dp[zy]);ans[i+1] = maxx;}for(int i = n;i>= 1;i--)printf("%lld\n",ans[i]);return 0;}

AC代码②：

#include<cstdio>//网上普遍写法#include<iostream>#include<algorithm>#include<queue>#include<stack>#include<cstring>#include<string>#include<vector>#include<cmath>#include<map>using namespace std;typedef long long ll;#define mem(a,b) memset(a,b,sizeof(a))const int maxn = 2e5+5;const int ff = 0x3f3f3f3f;int n;int ne[maxn];ll a[maxn],b[maxn];ll dp[maxn];ll ans[maxn];int find(int x){return ne[x] == x?x:ne[x] = find(ne[x]);}int main(){cin>>n;for(int i = 1;i<= n;i++){scanf("%I64d",&a[i]);ne[i] = i;}for(int i = n;i>= 1;i--)scanf("%I64d",&b[i]);for(int i = 1;i< n;i++){dp[b[i]] = a[b[i]];if(dp[b[i]-1]!= 0){int zy = find(b[i]-1);dp[b[i]] += dp[zy];ne[zy] = b[i];}if(dp[b[i]+1]!= 0){int zy = find(b[i]+1);dp[b[i]] += dp[zy];ne[zy] = b[i];}ans[i+1] = max(ans[i],dp[b[i]]);}ans[1] = 0;for(int i = n;i>= 1;i--)printf("%I64d\n",ans[i]);return 0;}