POJ-1042:Gone Fishing

시간 제한:
2000ms 
메모리 제한:
65536kB
묘사
John is going on a fishing trip. He has h hours available (1 <= h <= 16), and there are n lakes in the area (2 <= n <= 25) all reachable along a single, one-way road. John starts at lake 1, but he can finish at any lake he wants. He can only travel from one lake to the next one, but he does not have to stop at any lake unless he wishes to. For each i = 1,...,n - 1, the number of 5-minute intervals it takes to travel from lake i to lake i + 1 is denoted ti (0 < ti <=192). For example, t3 = 4 means that it takes 20 minutes to travel from lake 3 to lake 4. To help plan his fishing trip, John has gathered some information about the lakes. For each lake i, the number of fish expected to be caught in the initial 5 minutes, denoted fi( fi >= 0 ), is known. Each 5 minutes of fishing decreases the number of fish expected to be caught in the next 5-minute interval by a constant rate of di (di >= 0). If the number of fish expected to be caught in an interval is less than or equal to di , there will be no more fish left in the lake in the next interval. To simplify the planning, John assumes that no one else will be fishing at the lakes to affect the number of fish he expects to catch. 
Write a program to help John plan his fishing trip to maximize the number of fish expected to be caught. The number of minutes spent at each lake must be a multiple of 5.
입력
You will be given a number of cases in the input. Each case starts with a line containing n. This is followed by a line containing h. Next, there is a line of n integers specifying fi (1 <= i <=n), then a line of n integers di (1 <=i <=n), and finally, a line of n - 1 integers ti (1 <=i <=n - 1). Input is terminated by a case in which n = 0.
출력
For each test case, print the number of minutes spent at each lake, separated by commas, for the plan achieving the maximum number of fish expected to be caught (you should print the entire plan on one line even if it exceeds 80 characters). This is followed by a line containing the number of fish expected. 
If multiple plans exist, choose the one that spends as long as possible at lake 1, even if no fish are expected to be caught in some intervals. If there is still a tie, choose the one that spends as long as possible at lake 2, and so on. Insert a blank line between cases.
샘플 입력

2


1 


10 1 


2 5 


2 


4 


4 


10 15 20 17 


0 3 4 3 


1 2 3 


4 


4 


10 15 50 30 


0 3 4 3 


1 2 3 


0
 

샘플 출력

45, 5 


Number of fish expected: 31 





240, 0, 0, 0 


Number of fish expected: 480 





115, 10, 50, 35 


Number of fish expected: 724 

이 [매거] + [욕심]의 제목, 우선 첫 번째 생각의 점은 무엇을 매거하는가?이 문제는 모두 몇 개의 호수를 지나는지 일일이 열거한다.둘째, 욕심 뭐?이 문제는 호수마다 현재 상태에서 5분 동안 잡을 수 있는 물고기의 수를 욕심내면 첫 번째 5분은 1번 호수, 두 번째 5분은 13번 호수, 그러면 바로 13번 호수로 가야 하지 않겠는가?이것은 이 문제를 교묘하게 다루는 것이다. 이때 바로 13호 호수에 도착할 필요가 없고 기장을 채택하여 13호는 5분 동안 낚시를 해야 하며, 장래에 13호 호수에 도착할 때 집행해야 한다.
해법:

#include <iostream>
using namespace std;

void display(int *arr,int n)
{
	for(int i=0; i<n;++i)
	{	
		cout<<arr[i]*5;	
		if(i!=n-1)
			cout<<", ";
	}
	cout<<endl;
}

int max(int *arr,int n)
{
	int max_pos =0;
	for(int i=1; i<n; ++i)
	{	if(arr[max_pos]<arr[i])
			max_pos = i;
	}
	return max_pos;
}

int main()
{
	int n; cin>>n;
	while(n)
	{

		int h; cin>>h;
		int fsh[25] = {};
		int dec[25] = {}; 
		int trv[25] = {};
		for(int i=0; i<n; ++i)
		{	cin>>fsh[i];	}
		for(int i=0; i<n; ++i)
		{	cin>>dec[i];	}
		for(int i=0; i<n-1; ++i)
		{	cin>>trv[i];	}


		/*calculate the result*/
		int spend [25][25] = {};
		int fishes[25]    = {}; /* */

		/* ， 0 n-1 */
		for(int i=0;i<n;++i)
		{	
			/* i */
			int intervals = h*12;
			for(int j=0; j<i; ++j)
			{	intervals -= trv[j];}

			for(int j=0;j<intervals;++j)
			{	
				int max  = fsh[0]-spend[i][0]*dec[0];
				if(max<0)max = 0;
				int lake = 0;
				for(int j=0; j<=i; ++j)
				{	
					int tmp = fsh[j]-spend[i][j]*dec[j];
					if(tmp<0){	tmp=0;	}
					if(max < tmp)
					{	
						max  = tmp;
						lake = j;
					}
				}
				spend[i][lake]++;
				fishes[i] += max;
			}	
		}

	
		int lakes = max(fishes,n);
		display(spend[lakes],n);
		cout<<"Number of fish expected: "
			 <<fishes[lakes]<<endl<<endl;

		cin>>n;
	}
	return 0;
}