CodeForces 722C Destroying Array（数组删除区间最大值）

http://codeforces.com/contest/722/problem/C

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.

题意：

给定两组数字，第一组是数字的大小，第二组是删除的次序，问删除后的最大区间和

思路：

并查集反向建立关系图即可

ac code：

#include<stdio.h>
#include<cstring>
#include<algorithm>
#define AC main()
using namespace std;
typedef __int64 LL;
const int MYDD = 1103 + 1e5;

int Set[MYDD], pos[MYDD];
LL Ans[MYDD], a[MYDD], sum[MYDD];
bool used[MYDD];

void init(int n) {
for(int j = 0; j <= n; j++){
Set[j] = j;
used[j] = 0;//标记未使用
sum[j] = 0;//临时记录的和
}
}

int Find(int x) {
return x == Set[x] ? x:Find(Set[x]);
}

void Combine(int x, int y) {
int fx = Find(x), fy = Find(y);
if(fx != fy) {
Set[fx] = fy;
sum[fy] += sum[fx];
}
}

int AC {
//	while(1) {
int n;
scanf("%d", &n);
init(n);
for(int j = 1; j <= n; j++)
scanf("%I64d", &a[j]);
for(int j = 1; j <= n; j++)
scanf("%d", &pos[j]);

LL tmp = 0;
for(int j = n; j >= 1; j--) {
int Now = pos[j];
sum[Now] = a[Now];
used[Now] = 1;

if(Now < n && used[Now+1])
Combine(Now, Now+1);

if(Now > 1 && used[Now-1])
Combine(Now, Now-1);

Ans[j] = tmp;
tmp = max(tmp, sum[Find(Now)]);
}
for(int j = 1; j <= n; j++)
printf("%I64d\n", Ans[j]);
//	}
return 0;
}