624 - CD UVA

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

这是最基础的背包问题，特点是：每种物品仅有一件，可以选择放或不放。

用子问题定义状态：即f[i][v]表示前i件物品恰放入一个容量为v的背包可以获得的最大价值。则其状态转移方程便是：

f[i][v]=max{f[i-1][v],f[i-1][v-c[i]]+w[i]}

这个方程非常重要，基本上所有跟背包相关的问题的方程都是由它衍生出来的。所以有必要将它详细解释一下：“将前i件物品放入容量为v的背包中”这个子问题，若只考虑第i件物品的策略（放或不放），那么就可以转化为一个只牵扯前i-1件物品的问题。如果不放第i件物品，那么问题就转化为“前i-1件物品放入容量为v的背包中”，价值为f[i-1][v]；如果放第i件物品，那么问题就转化为“前i-1件物品放入剩下的容量为v-c[i]的背包中”，此时能获得的最大价值就是f[i-1][v-c[i]]再加上通过放入第i件物品获得的价值w[i]。

http://www.cnblogs.com/Yu2012/archive/2011/12/02/2272092.html

http://blog.youkuaiyun.com/shiqi_614/article/details/6902569

http://blog.youkuaiyun.com/aidayei/article/details/6766390

备份：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int t[30];
int c[30][1005];
int vis[30];
int main()
{
    int total_time,number_track;
    while(scanf("%d%d",&total_time,&number_track)==2)
    {
        memset(c,0,sizeof(c));
        memset(vis,0,sizeof(vis));
        for(int i = 1;i<=number_track;++i)
           scanf("%d",&t[i]);
        for(int i = 1;i<=number_track;++i)
          for(int j = 1;j<=total_time;++j)
          {
              if(t[i]>j)
              c[i][j] = c[i-1][j];
              else
              c[i][j]=max(c[i-1][j],c[i-1][j-t[i]]+t[i]);
          }
          int k = c[number_track][total_time];
         for(int i = 1;i<= number_track;++i)
           if(c[i][k]>c[i-1][k])
           {
            vis[i]=1;
            k  -= t[i];
           }
          int j = c[number_track][total_time];
          for(int i = 1;i<=number_track&&j>0;++i)
          if(vis[i])
          {
          printf("%d ",t[i]);
          j -= t[i];
          }

          printf("sum:%d\n",c[number_track][total_time]);
    }

}