uva624-CD（背包）-优快云博客

本文介绍了一种解决如何从CD中选择曲目以最优化填充固定长度录音带空间的问题。该问题被视为一种背包问题，通过动态规划算法来确定最佳曲目组合，使得录音带的空间利用率最高且未使用的空间尽可能短。

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题目：

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

粘贴的有点问题

Memory: 0 KB Time: 15 MS Language: C++ 4.5.3Result: Accepted

这个题我是当作背包问题做的，仔细分析后才发现的，很简单

#include<stdio.h>  
#include<iostream>  
using namespace std;   
int c[35],w[35],f[430000][30],visit[35];  
void print(int i,int j)  
{  
  
    if(!j)return ;  
    if(f[i][j] == f[i][j-1]) print(i, j-1);  
    else  
    {  
        print(i-c[j-1],j-1 );  
        printf("%d ", c[j-1]);  
    }  
  
}  
int main()  
{  
    int i,j,n,v,m;  
    while(scanf("%d%d",&m,&n)!=EOF)  
    {  
        for(i=0;i<=m;i++)  
        for( j=0;j<=n;j++)  
            f[i][j]=0;  
        for(i=0;i<n;i++)  
        {  
            scanf("%d",c+i);  
            w[i]=c[i];  
        }     
        for (i=1;i<=n;i++)   
        {  
            for (v=1; v <= m; v++)  
            {  
                if(v<c[i-1])  
                    f[v][i]=f[v][i-1];  
                if(v>=c[i-1])  
                f[v][i] = (f[v][i-1] > f[v - c[i-1]][i-1] + w[i-1]?f[v][i-1] : f[v - c[i-1]][i-1] + w[i-1]);  
            }  
          
        }  
      
        v=m;  
          
        print(m,n);  
            printf("sum:%d\n",f[m][n]);  
    }  
    return 0;  
}