动态规划(二维背包问题)：UVAoj 473

 Raucous Rockers 

You just inherited the rights to n previously unreleased songs recorded by the popular group Raucous Rockers. You plan to release a set of m compact disks with a selection of these songs. Each disk can hold a maximum of t minutes of music, and a song can not overlap from one disk to another. Since you are a classical music fan and have no way to judge the artistic merits of these songs, you decide on the following criteria for making the selection:

1. The songs will be recorded on the set of disks in the order of the dates they were written.
2. The total number of songs included will be maximized.

Input

The input consists of several datasets. The first line of the input indicates the number of datasets, then there is a blank line and the datasets separated by a blank line. Each dataset consists of a line containing the values of n, t and m (integer numbers) followed by a line containing a list of the length of n songs, ordered by the date they were written (Each is between 1 and t minutes long, both inclusive, and .)

Output

The output for each dataset consists of one integer indicating the number of songs that, following the above selection criteria will fit on m disks. Print a blank line between consecutive datasets.

Sample Input

2

10 5 3
3, 5, 1, 2, 3, 5, 4, 1, 1, 5

1 1 1
1

Sample Output

6

1

　　题意不清啊，

1. The songs will be recorded on the set of disks in the order of the dates they were written.

　　这句话没有卵用，你一看错就真错了。

　　

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 using namespace std;
 int n,m,t;
 int dp[][],tim[];

 int main()
 {
     int T;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d%d%d",&n,&t,&m);
         memset(dp,,sizeof(dp));
         for(int i=;i<=n;i++){
             if(i!=n)
                 scanf("%d,",&tim[i]);
             else
                 scanf("%d",&tim[i]);
         }

         for(int i=;i<=n;i++)
             for(int j=m;j>=;j--)
                 for(int k=t;k>=tim[i];k--)
                     dp[j][k]=max(dp[j][k],max(dp[j][k-tim[i]],dp[j-][t])+);

         printf("%d\n",dp[m][t]);
         if(T)printf("\n");
     }

     return ;
 }