CodeForces - 722C Destroying Array

CodeForces - 722C Destroying Array

有一丝小小的激动，这道题居然看题解看懂了，逆向思维，并查集。

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.

#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
int n;
ll a[100005];
int p[100005];
int f[100005];
ll sum[100005];
ll ans[100005];
bool vis[100005];
ll m=0;
int find(int x){
	return x==f[x]?x:find(f[x]);
}
void Union(int i,int j){
	i=find(i);
	j=find(j);
	sum[i]=sum[j]=sum[i]+sum[j];
	f[i]=j;
}

int main(){
	int i,j,id;
	memset(vis,0,sizeof(vis));
	memset(sum,0,sizeof(sum));
	scanf("%d",&n);
	for(i=1;i<=n;i++){
		scanf("%lld",&a[i]);
	}
	for(i=1;i<=n;i++){
		scanf("%d",&p[i]);
	}
	ans[n]=0;
	for(i=1;i<=n;i++){
		f[i]=i;
	}
	for(i=n;i>1;i--){
		id=p[i];
		sum[id]=a[id];
		if(vis[id-1]){
			Union(id-1,id);
		}
		if(vis[id+1]){
			Union(id+1,id);
		}
		if(sum[id]>m){
			m=sum[id];
		}
		vis[id]=1;
		ans[i-1]=m;
	}
	for(i=1;i<=n;i++){
		printf("%lld\n",ans[i]);
	}
	return 0;
}