*HDU3496 背包DP

时间：2022-07-28 01:57:13

Watch The Movie

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 7397    Accepted Submission(s): 2367

Problem Description
New
semester is coming, and DuoDuo has to go to school tomorrow. She
decides to have fun tonight and will be very busy after tonight. She
like watch cartoon very much. So she wants her uncle to buy some movies
and watch with her tonight. Her grandfather gave them L minutes to watch
the cartoon. After that they have to go to sleep.
DuoDuo list N
piece of movies from 1 to N. All of them are her favorite, and she wants
her uncle buy for her. She give a value Vi (Vi > 0) of the N piece
of movies. The higher value a movie gets shows that DuoDuo likes it
more. Each movie has a time Ti to play over. If a movie DuoDuo choice to
watch she won’t stop until it goes to end.
But there is a strange
problem, the shop just sell M piece of movies (not less or more then),
It is difficult for her uncle to make the decision. How to select M
piece of movies from N piece of DVDs that DuoDuo want to get the highest
value and the time they cost not more then L.
How clever you are! Please help DuoDuo’s uncle.
 
Input
The
first line of the input file contains a single integer t (1 ≤ t ≤ 10),
the number of test cases, followed by input data for each test case:
The first line is: N(N <= 100),M(M<=N),L(L <= 1000)
N: the number of DVD that DuoDuo want buy.
M: the number of DVD that the shop can sale.
L: the longest time that her grandfather allowed to watch.
The
second line to N+1 line, each line contain two numbers. The first
number is the time of the ith DVD, and the second number is the value of
ith DVD that DuoDuo rated.
 
Output
Contain one number. (It is less then 2^31.)
The total value that DuoDuo can get tonight.
If DuoDuo can’t watch all of the movies that her uncle had bought for her, please output 0.
 
Sample Input
1
3 2 10
11 100 1
2
9 1
 
Sample Output
3
 
Source
 
题意：
买电影，每一个电影有他的时长，价值，给出n个可以买的电影，只能从中买m个，并且电影的时长不超过l，问能够买到的最大价值。
代码：
 //二维费用背包，第一维l不用必须装满，第二维m必须装满，这就要特殊初始化，dp时如果只为负忽略。

 #include<iostream>

 #include<cstdio>

 #include<cstring>

 using namespace std;

 int t,n,m,l;

 int f[][];

 int main()

 {

     int tim,val;

     scanf("%d",&t);

     while(t--)

     {

         memset(f,-,sizeof(f));

         scanf("%d%d%d",&n,&m,&l);

         for(int i=;i<=l;i++)

         f[i][]=;             //初始化。

         for(int i=;i<=n;i++)

         {

             scanf("%d%d",&tim,&val);

             for(int j=l;j>=tim;j--)

             {

                 for(int k=m;k>=;k--)

                 {

                     if(f[j-tim][k-]<) continue;   //直为负跳过

                     f[j][k]=max(f[j][k],f[j-tim][k-]+val);

                 }

             }

         }

         if(f[l][m]<) printf("0\n");

         else printf("%d\n",f[l][m]);

     }

     return ;

 }

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