zoukankan      html  css  js  c++  java
  • The Best Peak Shape

    In many research areas, one important target of analyzing data is to find the best "peak shape" out of a huge amount of raw data full of noises. A "peak shape" of length L is an ordered sequence of L numbers { D1​​, ⋯, DL​​ } satisfying that there exists an index i (1) such that D1​​<<Di1​​<Di​​>Di+1​​>>DL​​.

    Now given N input numbers ordered by their indices, you may remove some of them to keep the rest of the numbers in a peak shape. The best peak shape is the longest sub-sequence that forms a peak shape. If there is a tie, then the most symmetric (meaning that the difference of the lengths of the increasing and the decreasing sub-sequences is minimized) one will be chosen.

    Input Specification:

    Each input file contains one test case. For each case, the first line gives an integer N (3). Then N integers are given in the next line, separated by spaces. All the integers are in [.

    Output Specification:

    For each case, print in a line the length of the best peak shape, the index (starts from 1) and the value of the peak number. If the solution does not exist, simply print "No peak shape" in a line. The judge's input guarantees the uniqueness of the output.

    Sample Input1:

    20
    1 3 0 8 5 -2 29 20 20 4 10 4 7 25 18 6 17 16 2 -1
    
     

    Sample Output1:

    10 14 25
    
     

    Sample Input2:

    5
    -1 3 8 10 20
    
     

    Sample Output2:

    No peak shape
     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 int main()
     4 {
     5 //    freopen("data.txt","r",stdin);
     6     int n,k,x;
     7     int msum=0,mmin=99999;
     8     scanf("%d",&n);
     9     vector<pair<int,int> > v(n,make_pair(0,0));
    10     vector<int> vm;
    11     for(auto& i:v)
    12     {
    13         scanf("%d",&x);
    14         i.first=x;
    15         if(vm.empty())
    16         vm.emplace_back(x);
    17         else
    18         {
    19             if(x>vm.back())
    20             {
    21                 vm.emplace_back(x);
    22                 i.second=vm.size()-1;
    23             }
    24             else
    25             for(int j=vm.size()-2;;--j)
    26             {
    27                 if(x>vm[j])
    28                 {
    29                     i.second=j+1;
    30                     if(x<vm[j+1])
    31                     vm[j+1]=x;
    32                     break;
    33                 }
    34                 if(j<0)
    35                 {
    36                     i.second=0;
    37                     if(x<vm[0])
    38                     vm[0]=x;
    39                     break;
    40                 }
    41             }
    42         }
    43     }
    44     vm.clear();
    45     x=0;
    46     for(int i=v.size()-1;i>-1;--i)
    47     {
    48         int p;
    49         if(vm.empty())
    50         {
    51             vm.emplace_back(v[i].first);
    52             p=0;
    53         }
    54         else
    55         {
    56             if(v[i].first>vm.back())
    57             {
    58                 vm.emplace_back(v[i].first);
    59                 p=vm.size()-1;
    60             }
    61             else
    62             for(int j=vm.size()-2;;--j)
    63             {
    64                 if(v[i].first>vm[j])
    65                 {
    66                     p=j+1;
    67                     if(v[i].first<vm[j+1])
    68                     vm[j+1]=v[i].first;
    69                     break;
    70                 }
    71                 if(j<0)
    72                 {
    73                     p=0;
    74                     if(v[i].first<vm[0])
    75                     vm[0]=v[i].first;
    76                     break;
    77                 }
    78             }
    79         }
    80         if(p+v[i].second>msum)
    81         {
    82             msum=p+v[i].second;
    83             mmin=abs(p-v[i].second);
    84             x=i;
    85         }
    86         else if(p+v[i].second==msum)
    87         {
    88             if(mmin>abs(p-v[i].second))
    89             {
    90                 mmin=abs(p-v[i].second);
    91                 x=i;
    92             }
    93         }
    94     }
    95     if(x==v.size()-1||x==0)
    96     printf("No peak shape");
    97     else
    98     printf("%d %d %d",msum+1,x+1,v[x].first);
    99 }
    诚者,君子之所守也。
  • 相关阅读:
    初试PL/SQL并行编程
    SVN 让项目某些文件不受版本控制
    分析php获取客户端ip
    一道月薪3W的java面试题 (小明和小强都是张老师的学生,张老师的生日是某月某日,2人都不知道张老师的生日)
    js遍历对象的属性并且动态添加属性
    CodeForces 150B- Quantity of Strings 推算..
    linux 中多线程使用
    Microsoft Deployment Toolkit 2013 Preview Release Now Available
    谁有SMI-S Provider的一些源码,能参考一下吗
    hdu1395-2^x mod n = 1
  • 原文地址:https://www.cnblogs.com/SkystarX/p/12285782.html
Copyright © 2011-2022 走看看