Intel Code Challenge Elimination Round (Div.1 + Div.2, combined) C 倒序并查集-优快云博客

本文介绍了一种使用并查集数据结构解决特定数组问题的方法。问题要求按指定顺序移除数组元素，并在每次移除后找出剩余部分的最大连续子数组之和。通过倒序处理输入，并利用并查集进行高效合并，最终实现了问题的有效解决。

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 1to 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

input

5
1 2 3 4 5
4 2 3 5 1

output

input

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

output

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 integers1 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：依次给你n个正数，然后给你n个操作，每次都将其中一个数字划去，然后问你整个数组中连续的一段的的最大的和是多少。

并查集：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <cmath>
#include <map>
#include <bitset>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <set>

#define MM(a,b) memset(a,b,sizeof(a));
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define CT continue
#define SC scanf
const int N=1e5+10;
ll a[N],ans[N];
int op[N],use[N],f[N];

int findr(int u)
{
    if(f[u]!=u)
        f[u]=findr(f[u]);
    return f[u];
}

int main()
{
     int n;
     while(~SC("%d",&n))
     {
         for(int i=1;i<=n;i++)
         {
            SC("%lld",&a[i]);
            f[i]=i;
         }
         for(int i=1;i<=n;i++) SC("%d",&op[i]);
         MM(use,0);MM(ans,0);
         for(int i=n-1;i>=1;i--)
         {
             int k=op[i+1];
             use[k]=1;
             if(use[k-1])
             {
                int par=findr(k-1);
                a[k]+=a[par];
                f[par]=k;
             }
             if(use[k+1])
             {
                int par=findr(k+1);
                a[k]+=a[par];
                f[par]=k;
             }
             ans[i]=max(a[k],ans[i+1]);
         }
         for(int i=1;i<=n;i++) printf("%lld\n",ans[i]);
     }
     return 0;
}

　　倒序并查集，合并。

转载于:https://www.cnblogs.com/smilesundream/p/5927328.html