学步园

Gone Fishing

描述

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.

输入

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.

输出

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.

样例输入

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 

样例输出

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 

来源

East Central North A

上传者

张云聪

//我的策略是使用 链表动态增长， -
-  好久不用linklist   debug 很久才修正了，  顺带吐槽这个输出  有空格 不说明

package GoFishing;

import java.util.Scanner;

//acm30
public class GoFishing {

    public
static void main(String[] args) {
   
    GoFishing
goFishing = new GoFishing();
   
   
goFishing.solution();
    }

    public void
solution() {
   
    in = new
Scanner(System.in);
   
   
while(getdata()){
   
   
   
initial();
   
   
   
handle();
   
   
   
ShowResult();
   
   
   
System.out.println();
   
    }
    }
    Scanner
in;
    int
size;
    int
time;
    int[]
lake;
    int[]
decreaseRATE;
    int[]
distance;

    //
获取数据
    private
boolean getdata() {
   
    size =
in.nextInt();
   
    if(size ==
0){
   
   
  return  false;
   
    }
   
    time =
in.nextInt() * 12; // 将小时为单位的时间转化为以5min 为单位

   
    lake = new
int[size ];
   
    for (int i =
0; i < size; i++) {
   
   
    lake[i] =
in.nextInt();
   
    }

   
    decreaseRATE
= new int[size];

   
    for (int i =
0; i < size; i++) {
   
   
   
decreaseRATE[i] = in.nextInt();
   
    }

   
    distance =
new int[size - 1];
   
    for (int i =
0; i < size - 1; i++) {
   
   
    distance[i]
= in.nextInt();
   
    }
   
    return
true;
    }

    //
处理数据
    private void
handle() {
   
    for(int i =
0 ; i < size ;i++){
   
   
   
add(i+1);