Codeforces 722C Destroying Array （并查集）

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
题目链接：http://codeforces.com/problemset/problem/722/C

题意：给定n个数，然后第二行也有n个数（按照输入的顺序表示第一行n个数被破坏的顺序），问求当破坏掉第i个数时，当前序列中最大的连续字段和为多少（连续子序列中不能包含被破坏的数字）。

思路：按照正向删除的思路想了好久都没有想出来合适的方法（除了n^2的暴力），真的应了一句话：正难则反。删除不好操作那就换成添加。按照删除的逆序添加这些元素，当添加第i个数字的时候，判断第i-1和第i+1个数字是否被添加进去即可。如果被添加进去，则是一个连续的子段，否则不是。最后求出一个最大的就可以啦。

【并查集】通过并查集来进行子段和的合并和查询。

代码如下：

#include <bits/stdc++.h>
#define LL long long
using namespace std;

const int N = 1e5+10;
int b
,n,par
,vis
;
LL ans
,a
;

int FindSet(int x){
if(x == par[x]) return x;
return par[x] = FindSet(par[x]);
}

int main(){
scanf("%d",&n);
for(int i = 1; i <= n; i ++) scanf("%I64d",&a[i]);
for(int i = 1; i <= n; i ++) scanf("%d",&b[i]);
for(int i = 1; i <= n; i ++) par[i] = i;
memset(vis,0,sizeof(vis));
ans
= 0;
for(int i = n; i > 0; i --){
int x = b[i];
if(vis[x-1]){
int fx = FindSet(x);
int fy = FindSet(x-1);
par[fy] = fx;
a[fx] += a[fy];
}
if(vis[x+1]){
int fx = FindSet(x);
int fy = FindSet(x+1);
par[fy] = fx;
a[fx] += a[fy];
}
ans[i-1] = max(ans[i],a[FindSet(x)]);
vis[x] = 1;
}
for(int i = 1; i <= n; i ++) printf("%I64d\n",ans[i]);
return 0;
}