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.

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題目大意：給出一個長度為n的序列，每次刪除一個（刪除之后序列斷開），求最大連續子段和。（序列中數為正整數）

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sol:正著做的話感覺做法有點諧，那么把詢問離線，用并查集將數字一個一個連通起來，每一個點帶一個權值，將每一個連續序列的權值記到父親節點上

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? 1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 #include<cstdlib> 5 #include<algorithm> 6 #include<vector> 7 #include<cmath> 8 #include<ctime> 9 #include<cstring> 10 #define yyj(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout); 11 #define llg long long 12 #define maxn 100010 13 #define md 50000 14 #define inf 0x7fffffff 15 using namespace std; 16 llg i,j,k,n,m,a[maxn],f1,f2,maxl,c[maxn],ans[maxn],dad[maxn],bj[maxn],x,val[maxn]; 17 18 llg find (llg x) 19 { 20 return dad[x]==x?x:dad[x]=find(dad[x]); 21 } 22 23 int main() 24 { 25 // yyj("c"); 26 cin>>n; 27 for (i=1;i<=n;i++) scanf("%I64d",&a[i]),dad[i]=i; 28 for (i=1;i<=n;i++) scanf("%I64d",&c[i]); 29 for (i=n;i>=1;i--) 30 { 31 maxl=max(maxl,a[c[i]]); 32 bj[c[i]]=1; val[c[i]]+=a[c[i]]; 33 x=c[i]; 34 if (bj[x-1]!=0 && bj[x+1]!=0) 35 { 36 f1=find(x-1); 37 dad[find(x)]=f1; 38 f2=find(x+1); 39 dad[f2]=f1; 40 val[f1]+=a[x]+val[x+1]; 41 maxl=max(maxl,val[f1]); 42 } 43 else 44 if (bj[x-1]!=0) 45 { 46 f1=find(x-1); 47 dad[find(x)]=f1; 48 val[f1]+=val[x]; 49 maxl=max(maxl,val[f1]); 50 } 51 else 52 if (bj[x+1]!=0) 53 { 54 f2=find(x+1); 55 dad[f2]=find(x); 56 val[find(x)]+=val[f2]; 57 maxl=max(maxl,val[find(x)]); 58 } 59 else 60 { 61 dad[x]=x; 62 val[x]=a[x]; 63 maxl=max(maxl,a[x]); 64 } 65 ans[i]=maxl; 66 } 67 for (i=2;i<=n;i++) cout<<ans[i]<<endl; 68 cout<<0; 69 return 0; 70 }

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轉載于:https://www.cnblogs.com/Dragon-Light/p/5927527.html

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