Codeforces Round 722C：Destroying Array(并查集,好题)

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博客详细解析了Codeforces Round 722C的问题，涉及如何销毁数组元素并找出未被销毁元素的最大和子数组。通过反向操作，利用并查集处理加入数值时的线段最大和，并采用离线处理策略优化算法。

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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：

起先你有n个正整数a1,a2,...,an，现在你要将这个数组里的数一个接一个的破坏掉，直到最后每个数都被
破坏，对于每次破坏，求出当前最大连续段和，要求这些段不能包含被破坏的数

题解：

正着删除不好处理，就反过来求了，相当于从一个空序列往上面添加数值，加一次，求一次线段的最大和，使用离线处理，每次与当前值和前一个答案来更新答案！

加入某个值后，使用用并查集进行处理，令a[i]是 i 这个点的权值，那么先判断i+1的位置，如果有值的话，就把i+1这一系列线段与i位置上的值合并，再判断i-1位置。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
using namespace std;
typedef long long ll;
const int MAXN=1e5+100;
ll a[MAXN],ans[MAXN],f[MAXN];
int p[MAXN],pa[MAXN];
bool have[MAXN];                //记录某个点是否被添加 
int n;
void init()
{
	for(int i=1;i<=n;i++) pa[i]=i;
	memset(have,false,sizeof(false));
}
int find(int x)
{
	return pa[x]==x?x:pa[x]=find(pa[x]);
}
int main()
{
	scanf("%d",&n);
	init();
	for(int i=1;i<=n;i++) scanf("%I64d",&a[i]);
	for(int i=1;i<=n;i++) scanf("%d",&p[i]);
	ans[n]=0;
	for(int i=n;i>1;i--)
	{
		int p1=p[i];
		f[p1]=a[p1];
		if(have[p1-1])
		{
			int f1=find(p1),f2=find(p1-1);
			pa[f1]=f2;
			f[f2]+=f[f1];
		}
		if(have[p1+1])
		{
			int f1=find(p1),f2=find(p1+1);
			pa[f1]=f2;
			f[f2]+=f[f1];
		}
		ans[i-1]=max(f[find(p1)],ans[i]);
		have[p1]=true;
	}
	for(int i=1;i<=n;i++)
	printf("%I64d\n",ans[i]);
	return 0;
}