SOJ 1046

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 :” where is the index of the run. 

* One line for every requested period: “-” where is first quarter and 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

思路比较简单，只需要将所有的可能列举出来，然后再按题目要求的优先级排序，然后依次输出就行了。

#include
#include
#include
#include
#include
using namespace std;

int N,quarter_number,request_number,least_number;//N为case数，quarter_number为总的时间，request_number为要求输出的方法数量，least_number为每次spotting最短的时长 
int planes_per_quarter[301];//用来存储每刻钟的飞机数量 
struct node{
	int start;//起时刻 
	int end;//终止时刻 
	int length;//时长，即从开始到终止共几刻钟 
	double average;//平均每刻钟的飞机数量 
	node(int s,int e)
	{
		start=s;
		end=e;
		length=e-s+1;
		double sum=0;
		for(int i=s;i<=e;++i)
		{
			sum+=planes_per_quarter[i];
		}
		average=sum/length;
	}
};

vector list;//list容器用来存储所有的可行方案 
void saveData()//读取每刻钟飞机的数量并存入数组中 
{
	int number;
	for(int i=0;i>number;
		planes_per_quarter[i+1]=number;
	}
}
void make_candidate()//穷举所有可能的方法并存入list容器 
{
	list.clear();
	for(int i=1;i<=quarter_number-least_number+1;++i)//注意两个循环的条件，要使方法满足时长不少于 least_number
	{
		for(int j=i+least_number-1;j<=quarter_number;++j)
		{
			list.push_back(node(i,j));
		}
	}
}
bool cmp(node A,node B)//排序的判断函数，按照题目要求的优先级来即可 
{
	if(A.average!=B.average) return A.average>B.average;
	else if(A.length!=B.length) return A.length>B.length;
	else return A.end>N;
	int count=1;
	for(int i=0;i>quarter_number>>request_number>>least_number;
		saveData();
		make_candidate();
		sort(list.begin(),list.end(),cmp);
		cout<<"Result for run "<