1046. Plane Spotting

1046. Plane Spotting

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

Craig is fond of planes. Making photographs of planes forms a major part of his daily life. Since he tries to stimulate his social life, and since it’s quite a drive from his home to the airport, Craig tries to be very efficient by investigating what the optimal times are for his plane spotting. Together with some friends he has collected statistics of the number of passing planes in consecutive periods of fifteen minutes (which for obvious reasons we shall call ‘quarters’). In order to plan his trips as efficiently as possible, he is interested in the average number of planes over a certain time period. This way he will get the best return for the time invested. Furthermore, in order to plan his trips with his other activities, he wants to have a list of possible time periods to choose from. These time periods must be ordered such that the most preferable time period is at the top, followed by the next preferable time period, etc. etc. The following rules define which is the order between time periods:

1. A period has to consist of at least a certain number of quarters, since Craig will not drive three hours to be there for just one measly quarter.
2. A period P1 is better than another period P2 if:
* the number of planes per quarter in P1 is higher than in P2;
* the numbers are equal but P1 is a longer period (more quarters);
* the numbers are equal and they are equally long, but period P1 ends earlier.

Now Craig is not a clever programmer, so he needs someone who will write the good stuff: that means you. So, given input consisting of the number of planes per quarter and the requested number of periods, you will calculate the requested list of optimal periods. If not enough time periods exist which meet requirement 1, you should give only the allowed time periods.

Input

The input starts with a line containing the number of runs N. Next follows two lines for each run. The first line contains three numbers: the number of quarters (1–300), the number of requested best periods (1–100) and the minimum number of quarters Craig wants to spend spotting planes (1–300). The sec-nod line contains one number per quarter, describing for each quarter the observed number of planes. The airport can handle a maximum of 200 planes per quarter.

Output

The output contains the following results for every run: 

* A line containing the text “Result for run <N>:” where <N> is the index of the run. 

* One line for every requested period: “<F>-<L>” where <F> is first quarter and <L> is the last quarter of the period. The numbering of quarters starts at 1. The output must be ordered such that the most preferable period is at the top.

Sample Input

3
10 5 5
1 5 0 2 1 4 2 5 0 2 
10 3 5
10 3 1 4 2 6 3 0 8 0 
5 5 5
1 2 3 4 5

Sample Output

Result for run 1:
4-8
2-8
6-10
1-8
2-6
Result for run 2:
1-6
1-7
1-9
Result for run 3:
1-5
题解：第一行为测试用例个数，第二行三个数分别为quarter个数，最优解个数，最小quarter个数，接下来一行为每一个quarter的飞机架数，题目要求解出符合题中给出三个要求的最优解，即对所有可能的情况按照给出的规则进行排序，取出前面n个即可，不知道为何用快排过不了。。。代码如下，sort可以测试通过，qsort不行，谁能帮忙解答？？？

// Problem#: 1046
// Submission#: 2209677
// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License
// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/
// All Copyright reserved by Informatic Lab of Sun Yat-sen University
#include <iostream>
#include <algorithm>

using namespace std;

struct answer{
       int start;
       int end;
       double avarage;
};

/*
int compare(const void* a, const void* b){
    int num_quarters_a = (*(answer*)a).end - (*(answer*)a).start;
    int num_quarters_b = (*(answer*)b).end - (*(answer*)b).start;
    if((*(answer*)a).avarage < (*(answer*)b).avarage)
         return 1;
    else if((*(answer*)a).avarage > (*(answer*)b).avarage)
         return -1;
    else{
         if(num_quarters_a < num_quarters_b)
              return 1;
         else if(num_quarters_a > num_quarters_b)
              return -1;
         else if((*(answer*)a).end > (*(answer*)b).end)
             return 1;
         return -1;
    }
}
*/
bool compare(answer a, answer b){
     int num_quarters_a = a.end - a.start;
     int num_quarters_b = b.end - b.start;
     if(a.avarage > b.avarage)
          return true;
     else if(a.avarage < b.avarage)
          return false;
     else{
          if(num_quarters_a > num_quarters_b)
              return true;
          else if(num_quarters_a > num_quarters_b)
              return false;
          else if(a.end < b.end)
             return true;
          return false;      
     }
}

int main(){
    int case_number;
    cin>>case_number;
    int runs = 1;
    while(runs <= case_number){
         int quarters;
         int best_periods;
         int min_num_quarters;
         cin>>quarters>>best_periods>>min_num_quarters;
         int* quarters_detail = new int[quarters];
         answer answer_hold[100000]; 
         
         //int n = (quarters+2-min_num_quarters)*((quarters-min_num_quarters+1)/2);
         for(int i = 0; i < quarters; i++){
                 cin>>quarters_detail[i];
         }
         
         int hold_count = 0;
         for(int k = min_num_quarters; k <= quarters; k++){
             for(int i = 0; i <= quarters-k; i++){
                     int sum = 0;
                     for(int j = i; j < k+i; j++){
                             sum += quarters_detail[j];
                     }
                     answer_hold[hold_count].start = i+1;
                     answer_hold[hold_count].end = i+k;
                     answer_hold[hold_count].avarage = (double)sum/k;
                     hold_count++;
             }
         }
         
         sort(answer_hold,answer_hold+hold_count,compare);
         
         cout<<"Result for run "<<runs<<":"<<endl;
         
         for(int i = 0; i < best_periods && i < hold_count; i++){
                 cout<<answer_hold[i].start<<"-"<<answer_hold[i].end<<endl;
         }
         runs++;
    }
    return 0;    
}