Intel Code Challenge Elimination Round (Div.1 + Div.2, combined) C 并查集

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C. 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.

题意：

给出长度为n的>=0的整数序列val[i]

接下来给出长度为n的损坏序列b[i]

表示当前将b[i]这个位置的val[i]去除，然后求剩下的连续的区间内和的最大值

每次去除一个位置的权值，输出当前k个区间的最大区间和

思路：

一开始模拟过程的时候想着是当前去除一个，剩下的断开的区间如何快速找出最大的区间和

所以肯定也会逆向思维一下，从n->1慢慢的将区间补齐，那么当前位置i插入一个数，利用并查集的合并功能，就可以到达合并区间的效果

1：当前位置i左边已经有数插入了且右边已经有数插入了，则快速合并左右两区间为一区间然后更新大区间的区间和

2：当前位置i左边已经有数插入了

3：当前位置i右边已经有数插入了

4：当前位置i左右都没有插入数

代码：
#include <iostream>
using namespace std;
int vis[100100];
int fa[100100];
int n;
int a[100100],b[100100];
long long anw[100100],sum[100100];
int Find(int x){
if(x!=fa[x]) fa[x]=Find(fa[x]);
return fa[x];
}
int main(){
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=0;i<=n+1;i++) fa[i]=i;
long long ans=0;
for(int i=n;i>=1;i--){
if(vis[b[i]+1] && vis[b[i]-1]){
int X=Find(b[i]+1);
int Y=Find(b[i]-1);
///printf("%d %d\n",X,Y);
fa[X]=Y;
fa[b[i]]=Y;
sum[Y]+=1LL*(a[b[i]]+sum[X]);
ans=max(ans,sum[Y]);
} else if(vis[b[i]+1]){
int X=Find(b[i]+1);
fa[b[i]]=X;
sum[X]+=a[b[i]];
ans=max(ans,sum[X]);
} else if(vis[b[i]-1]){
int X=Find(b[i]-1);
fa[b[i]]=X;
sum[X]+=a[b[i]];
ans=max(ans,sum[X]);
} else {
int X=Find(b[i]);
sum[X]+=a[b[i]];
ans=max(ans,sum[X]);
}
vis[b[i]]=1;
anw[i]=ans;
}
for(int i=2;i<=n;i++) cout<<anw[i]<<endl;
cout<<0<<endl;
}