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  • Codeforces 722C. Destroying Array

    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:

    1. Third element is destroyed. Array is now 1 3  *  5. Segment with maximum sum 5 consists of one integer 5.
    2. Fourth element is destroyed. Array is now 1 3  *   * . Segment with maximum sum 4 consists of two integers 1 3.
    3. First element is destroyed. Array is now  *  3  *   * . Segment with maximum sum 3 consists of one integer 3.
    4. Last element is destroyed. At this moment there are no valid nonempty segments left in this array, so the answer is equal to 0.

    题目大意:给出一个长度为n的序列,每次删除一个(删除之后序列断开),求最大连续子段和。(序列中数为正整数)



     
     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 }
    本文作者:xrdog 作者博客:http://www.cnblogs.com/Dragon-Light/ 转载请注明出处,侵权必究,保留最终解释权!
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  • 原文地址:https://www.cnblogs.com/Dragon-Light/p/5927527.html
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