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最新推荐文章于 2024-12-18 13:00:00 发布

转载最新推荐文章于 2024-12-18 13:00:00 发布 · 72 阅读

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原文链接：http://www.cnblogs.com/Tinamei/p/4732438.html

本文介绍了一种解决如何从CD中选择曲目以最优化填满固定时长磁带的问题。通过0-1背包问题算法及路径回溯，实现曲目的最佳组合，确保磁带空间得到充分利用且剩余空间最短。

You have a long drive by car ahead. You have a tape recorder, but unfortunately your best music is on CDs. You need to have it on tapes so the problem to solve is: you have a tape N minutes long. How to choose tracks from CD to get most out of tape space and have as short unused space as possible.

Assumptions:

number of tracks on the CD. does not exceed 20
no track is longer than N minutes
tracks do not repeat
length of each track is expressed as an integer number
N is also integer

Program should find the set of tracks which fills the tape best and print it in the same sequence as the tracks are stored on the CD

Input

Any number of lines. Each one contains value N, (after space) number of tracks and durations of the tracks. For example from first line in sample data: N=5, number of tracks=3, first track lasts for 1 minute, second one 3 minutes, next one 4 minutes

Output

Set of tracks (and durations) which are the correct solutions and string ``sum:" and sum of duration times.

Sample Input

5 3 1 3 4
10 4 9 8 4 2
20 4 10 5 7 4
90 8 10 23 1 2 3 4 5 7
45 8 4 10 44 43 12 9 8 2

Sample Output

1 4 sum:5
8 2 sum:10
10 5 4 sum:19
10 23 1 2 3 4 5 7 sum:55
4 10 12 9 8 2 sum:45

Miguel A. Revilla
2000-01-10

要经历N分钟长的时间，如何选择带子可以尽可能的利用时间

每个CD的时间不超过 20
没有哪个CD的时间是超过N的
CD不能重复
每个长度和N都是一个整数

数据：
一个N代表时间， M代表有M个CD

分析：
其实就是0-1背包+打印路径

 1 #include <iostream>
 2 #include <cstdlib>
 3 #include <cstdio>
 4 #include <algorithm>
 5 #include <vector>
 6 #include <queue>
 7 #include <cmath>
 8 #include <stack>
 9 #include <cstring>
10 
11 using namespace std;
12 
13 #define INF 0xfffffff
14 #define maxn 2000
15 #define min(a,b) (a<b?a:b)
16 #define max(a,b) (a>b?a:b)
17 int w[maxn];
18 int dp[maxn][maxn];
19 
20 void Path(int m,int n)   //  路径
21 {
22     if( m == 0)
23         return ;
24 
25     if(dp[m][n] == dp[m-1][n])
26         Path(m-1, n);
27     else
28     {
29         Path(m-1, n-w[m]);
30         printf("%d ",w[m]);
31     }
32 }
33 int main()
34 {
35     int n, m;
36 
37     while(cin >> n >> m)
38     {
39         memset(dp,0,sizeof(dp));
40 
41         for(int i = 1; i<=m; i++)
42             cin >> w[i];
43 
44         for(int i = 1; i<=m; i++)
45         {
46             for(int j=0; j<=n; j++)
47             {
48                 if(j<w[i])
49                     dp[i][j] = dp[i-1][j];
50                 else
51                     dp[i][j] = max(dp[i-1][j], dp[i-1][j-w[i]]+w[i]);  // 
52             }
53         }
54         Path(m,n);
55 
56         printf("sum:%d\n", dp[m][n]);
57     }
58     return 0;
59 }