-1055 group photo PAT B (25 minutes)

Click on the link full solution summary PAT B -AC

Title:
formation is very important when a group photo shoot, where formation of the following design rules given queuing K N individual row:

• The number of each row of N / K (rounded down), all the extra people standing in the last row;
• Back all child not shorter than anyone in the front row;
• Each intermediate station the highest row (middle position m / 2 + 1, where m for the number of rows, the division rounded down);
• In each row others intermediary axis, according to a non-increasing order of height, the first left and right sides alternately enqueue standing intermediary (e.g. 190,188,186,175,170 height of 5, the formation 175, . 188,190,186,170 this assumes you are facing the photographer, so your left is the right of intermediaries);
• If more than the same height, press the name of lexicographic ascending order. Here to ensure that no duplicate names.

Now given a set of pictures of people, write a program output their formation.

Input format:
Each input comprises a test. Each test case is given row of the first two positive integers N (≦ 10 ^{. 4} , the total number) and K (≤10, total number of rows). Then N rows, each row gives the name of a person (not including spaces, length of no more than 8 letters) and height ([30, 300] integer in the interval).

Output formats:
Output to take pictures of the formation. That K row names, separated by spaces between them, the end of the line may not have the extra space. Note: if you face the photographer, who back in the top of the output, the output of the front row at the bottom.

Sample input:

10 3
Tom 188
Mike 170
Eva 168
Tim 160
Joe 190
Ann 168
Bob 175
Nick 186
Amy 160
John 159

Sample output:

Bob Tom Joe Nick
Ann Mike Eva
Tim Amy John

My code:

#include<iostream>
#include<cstdio>
#include<vector>
#include<string>
#include<set>
#include<map>
#include<algorithm>
#include<cmath>
#include<ctime>
#include<cstring>
#include<sstream>
using namespace std;
//有的时候题目是一起做的，所以会有不需要的头文件

int cmp(pair<string,int>a,pair<string,int>b){
    if(a.second!=b.second)   return a.second>b.second;
    else return a.first<b.first;
}

void print_first_to_last(pair<string,int>student[],int first,int last)
{
    int m=last-first+1;
    int first_=(m%2==0)?(last):(last-1);
    int end_=(m%2==0)?(last-1):(last);

    int i=first_;
    while(i>=first+1)
    {
        cout<<student[i].first<<" ";
        i-=2;
    }
    i=first;
    while(1)
    {
        cout<<student[i].first;
        i+=2;
        if(i<=end_) cout<<" ";
        else
        {
            cout<<endl;
            return;
        }
    }
}


int main()
{
    int N,K;
    cin>>N>>K;
    pair<string,int>student[N];
    for(int i=0;i<N;i++)
    {
        string name;
        int height;
        cin>>name>>height;
        student[i]=make_pair(name,height);
    }
    sort(student,student+N,cmp);

    int num_line=N/K;
    print_first_to_last(student,0,num_line+N%K-1);
    for(int line=2;line<=K;line++)
        print_first_to_last(student, N%K+(line-1)*num_line, N%K + line*num_line -1);

    return 0;
}

My thoughts:

Determining the range of each row: A total of N individuals, in addition to the first line, there is K Line

1. First line of N/K+N%Kpeople

   The subscript 0~N/K+N%K-1

2. Then 2 ~ K lines each have N/Kpeople

   For the first linerow,
   the first index is a N/K + N%K + (line-2)*N/K= N%K + (line-1)*N/K
   the last index N%K + (line-1)*N/K + N/K-1=N%K + line*N/K -1

For the output of a row: Index for sorting, then according to the following target sequence one by one like the output

1. Sorting according to height, corresponding to subscript 0~N-1
2. For example, a line mnumber then there are two cases

• m is an even number: (m-1) ... 3 1 0 2 4 ... (m-2)

  • The first is: the last one
  • An intermediate stop at: 1 (after a first one)
  • The last one is: the last one before a

• m is odd: (m-2) ... 3 1 0 2 4 ... (m-3) (m-1)

  • The first is: the last one before a
  • An intermediate stop at: 1 (after a first one)
  • The last one is: the last one

3. For each row, one of the first index first, a last index last,m=last-first+1

• m is an even number: from last-2 -2 to first+1, and from first+2 to +2last-1
• m is odd: from last-1-2 -2 to first+1, and from first+2 to +2last