UVa 624 CD（01背包需要记录最优解的任一方案）

本文链接：https://blog.youkuaiyun.com/acm_code/article/details/25791111

在驾车长途旅行前，你需要将CD上的音乐转录到磁带上。磁带长度为N分钟，且每首歌长度不超过N分钟，不重复。目标是找到最佳的歌曲组合，使磁带空间利用率最高，剩余的未使用空间尽可能短。问题要求找到最优化的歌曲集合，并按CD上的原始顺序输出。输入包含CD上歌曲的数量和长度，程序需找出占用磁带空间最佳的歌曲序列。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

UVa 624 CD

链接：UVA 624

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

题意：CD上有n首歌，每首歌曲时长已知，歌曲没有重复的，想要将CD上的歌拷到磁带上，磁带的容量已知，求解拷贝哪几首歌会使磁带的空间使用率最大。（输出任意一种最优解的方案即可）

这题也是01背包问题，最优解很好求，但难点在于最优解的方案也要求出。因此在用一个二维数组p[i][v]标记容量为v时第i首歌是否选取。

AC代码：

#include<stdio.h>
#include<string.h>
int main()
{
    int a[25],p[25][10005],i,j,n,m,s[10005];
    while(scanf("%d%d",&m,&n)!=EOF){
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);
        memset(s,0,sizeof(s));
        memset(p,0,sizeof(p));
        for(i=1;i<=n;i++)
            for(j=m;j>=a[i];j--)
                if(s[j-a[i]]+a[i]>s[j]){
                    s[j]=s[j-a[i]]+a[i];
                    p[i][j]=1;
                }
        j=m;
        for(i=n;i>=1;i--)
            if(p[i][j]){
                printf("%d ",a[i]);
                j-=a[i];
            }
        printf("sum:%d\n",s[m]);
    }
    return 0;
}