1 second
256 megabytes
standard input
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.
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.
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.
4
1 3 2 5
3 4 1 2
5
4
3
0
5
1 2 3 4 5
4 2 3 5 1
6
5
5
1
0
8
5 5 4 4 6 6 5 5
5 2 8 7 1 3 4 6
18
16
11
8
8
6
6
0
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：正整数序列，每次毁灭一个元素，每次求剩下元素不包含毁灭元素的线段的最大和
yy了各种方法，BIT，线段树，甚至各种STL，然而.....
//
// main.cpp
// c
//
// Created by Candy on 10/1/16.
// Copyright © 2016 Candy. All rights reserved.
//
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long ll;
const int N=1e5+;
inline int read(){
char c=getchar();int x=,f=;
while(c<''||c>''){if(c=='-')f=-;c=getchar();}
while(c>=''&&c<=''){x=x*+c-'';c=getchar();}
return x;
}
int n,a[N],p[N],vis[N];
ll sum[N],mx=,st[N],top=;
int fa[N];
inline int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);} int main(){
n=read();
for(int i=;i<=n;i++) a[i]=read(),fa[i]=i;
for(int i=;i<=n;i++) p[i]=read();
for(int i=n;i>=;i--){
st[++top]=mx;
int cur=p[i],f1=,f2=;
vis[cur]=; sum[cur]+=a[cur];
if(vis[cur-]){f1=find(cur-);sum[cur]+=sum[f1];fa[f1]=cur;}
if(vis[cur+]){f2=find(cur+);sum[cur]+=sum[f2];fa[f2]=find(cur);}
mx=max(mx,sum[cur]);
}
while(top) printf("%I64d\n",st[top--]);
}