Poj/OpenJudge 1042 Gone Fishing

1.链接地址：

http://bailian.openjudge.cn/practice/1042/

http://poj.org/problem?id=1042

2.题目：

Gone Fishing

Time Limit: 2000MS		Memory Limit: 32768K
Total Submissions: 27652		Accepted: 8199

Description

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

Source

East Central North America 1999

3.思路：

DP

4.代码：

 //2010-04-22
 //v0.1 create by wuzhihui
 #include<iostream>
 using namespace std;

 int ti[];//ti (0 < ti <=192
 int fi[];
 int fi2[];
 int di[];
 int a[];
 int maxa[];
 int main()
 {
     int i,j;
     int h,h2;//hour between 1<=h<=16
     int n;//lake 2 <= n <= 25
     int fishes;
     int maxfishes,temp;
     int max;
     while((cin>>n)&&n!=)
     {
         cin>>h;
         for(i=;i<n;i++) cin>>fi[i];
         for(i=;i<n;i++) cin>>di[i];
         for(i=;i<n-;i++) cin>>ti[i];
         max=-;
         for(j=;j<n;j++) a[j]=;
         for(i=;i<n;i++)
         {
             h2=h*;
             //memset(a,0,sizeof(a));
             for(j=;j<n;j++) a[j]=;
             fishes=;
             for(j=;j<n;j++) fi2[j]=fi[j];
             for(j=;j<=i-;j++) h2-=ti[j];
             while(h2>)
             {
                 maxfishes=-;
                 for(j=;j<=i;j++)
                 {
                     if(fi2[j]>maxfishes)
                     {
                         maxfishes=fi2[j];
                         temp=j;
                     }//if
                 }//for_j
                 fishes+=maxfishes;
                 a[temp]++;
                 if(fi2[temp]<di[temp]) fi2[temp]=;
                 else fi2[temp]-=di[temp];
                 h2--;
             }//while_h2
             if(max<fishes)
             {
                 max=fishes;
                 for(j=;j<n;j++) maxa[j]=a[j];
             }//if
         }//while_n
         cout<<(maxa[]*);
         for(i=;i<n;i++) cout<<", "<<(maxa[i]*);
         cout<<endl;
         cout<<"Number of fish expected: "<<max<<endl<<endl;
     }

     return ;
 }