题解：Gone Fishing（枚举+贪心）

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.
Input
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.
Output
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.
Sample Input
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
Sample Output
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
题解：
我们可以把总时间分为两个部分：在路上的时间和在钓鱼的时间。由于路是单行的，所以在路上的时间取决于走的最远距离，即到达的池塘的最大编号。 剩余的时间用于钓鱼。我们可以把移动+钓鱼的混合过程拆分为移动的过程和钓鱼的过程，即指定一个池塘为终点，移动过程即从起点到终点的过程，钓鱼的过程就是从起点到终点的各个池塘中选择池塘钓鱼，因为移动过程所需的时间已经在前面考虑过了，这个时候我们就可以认为需要移动的时候可以直接瞬间到达。然后，选择到哪些池塘钓鱼的策略采用贪心的方法，每个钓鱼的5分钟都选择期望最大的那一个池塘，每在选择一个池塘钓鱼5分钟，减少相应池塘的期望，增加计划中在该池塘钓鱼的时间，然后继续选择期望最大的池塘，直到钓鱼的时间不够，或者池塘里没有鱼了。如果池塘没有鱼了，还有时间的话，把多余的时间分配给第一个池塘。

#include 
#include 
#include 
using namespace std;
const int N=25;
int n,h;
int f[N],d[N],t[N];//f第一个五分钟钓的鱼量，d为每个五分钟减少的鱼量，t为i到i+1五分钟的个数 
int ans;
int each[N];//记录最终每条湖用的时间
int tans,teach[N];//最优钓鱼量和各湖钓鱼时间 
int th,tf[N];//有效钓鱼时间和每条湖前五分钟的钓鱼量 
int main()
{
	int i,j;
	while(cin>>n&&n>0)
	{//当湖的数量为0的时候结束 
		cin>>h;//输入时间 
		for(i=0;i>f[i];//第一次的鱼量 
		for(i=0;i>d[i];//每五分钟减少的鱼量 
		for(i=0;i>t[i];//每个湖间距离需要的时间片 
		h*=12;//一小时12个时间片
		ans=-1;
		for(i=0;i0)
			{//当有效时间大于0 
				int ind=0,max=tf[0];//令第一条湖的鱼量为最大值 ，ind标记湖是第几条湖 
				for(j=0;j<=i;j++)
				{
					if(tf[j]>max)
					{//不考虑顺序先找第一次鱼量最大的湖 
						max=tf[j];
						ind=j;
					}
				}
				if(max==0)
				{//最大钓鱼量为0时，将剩余的钓鱼时间加到第一个湖上的钓鱼时间 
					teach[0]+=th*5;//例如样例一 
					break;
				}
				else
				{
					teach[ind]+=5;//最大湖的钓鱼时间，每钓一次加一次五 
					tans+=tf[ind];//加上最大鱼量的湖的该次的鱼数 
					tf[ind]=max(tf[ind]-d[ind],0);//如果鱼量不少于减少的鱼数 ，则减,小于减少数则赋值为0 
				} 
				th--;//有效时间减少一个时间片(一个时间片五分钟) 
			}
			if(i!=n-1)//i的话是表示在第i条湖停下来 
				h-=t[i];//减去到下一条湖的时间片 
			if(tans>ans)
			{//如果值大于前面的值，就把值赋给ans 
				ans=tans;
				for(j=0;j