UVa 624 CD

  CD 

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

 

给出一些歌曲的时长，给出一个时长为N的磁带，将这些歌曲中的一部分录制到磁带上，求最长可以录多长时间的歌（即使磁带的利用率最高），并列出所录的歌曲。

求最长时间的方法与UVa 562(http://www.cnblogs.com/lzj-0218/p/3565968.html）思路相同，用一个dp数组，dp[i]=true表示可以录出总长为i的歌，dp[i]=false表示不能。然后对每首歌track[j]，有dp[i]=dp[i-track[j]]

列出歌曲的方法就是再加一个数组pos，若dp[i-track[j]]=true，那么可以由状态转移方程推出dp[i]=true，这时就记录下pos[i]=j，全部算完之后，倒着推回去即可

 

#include<iostream>
#include<cstdio>
#include<stack>
#include<cstring>

using namespace std;

int n,number;
int track[30],pos[1000000];
bool dp[1000000];

int main()
{
    while(scanf("%d",&n)==1)
    {
        scanf("%d",&number);
        for(int i=0;i<number;i++)
            scanf("%d",&track[i]);

        memset(pos,-1,sizeof(pos));
        memset(dp,false,sizeof(dp));
        dp[0]=true;

        for(int i=0;i<number;i++)
        {
            for(int j=n;j>=track[i];j--)
                if(dp[j-track[i]]&&!dp[j])
                {
                    dp[j]=true;
                    pos[j]=i;
                }
        }

        int sum;
        stack<int> s;

        for(int i=n;i>=0;i--)
            if(dp[i])
            {
                sum=i;
                break;
            }

        int p=pos[sum],temp=sum;
        while(p!=-1)
        {
            s.push(track[p]);
            temp-=track[p];
            p=pos[temp];
        }

        while(!s.empty())
        {
            printf("%d ",s.top());
            s.pop();
        }
        printf("sum:%d\n",sum);
    }

    return 0;
}