Watch The Movie

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 4722 Accepted Submission(s): 1497

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

3 2 10

11 100

1 2

9 1

Sample Output

　　题意：给你n部电影，l分钟，每部电影有时长和观看价值，要你从中选不多不少m部电影出来，，保证在l分钟以内观看价值是最大的。

　　这是一条二维01背包，基本做法只需在一维的基础上再加一维，不过这里还有一个限制条件，那就是不多不少选m部，所以需要保证最终选到的电影恰好有m部，而不是最多有m部。

　　状态转移方程： j>=c[i] && k>=1 && dp[j-c[i]][k-1]>=0(意思是中间选了k-1部片子)时 dp[j][k] = max{dp[j][k] , dp[j-c[i]][k-1]+v[i]}

　　具体过程看代码

上代码：

#include <cstdio>
#include <cstring>
#define MAX 1001
#define max(x,y) (x > y ? x : y)
using namespace std;

int v[100],c[100],dp[MAX][101];
int n,m,l;

int main()
{
    int t;
    //freopen("data.txt","r",stdin);
    scanf("%d",&t);
    while(t--){
        scanf("%d %d %d",&n,&m,&l);
        for(int i=0;i<n;i++){
            scanf("%d %d",&c[i],&v[i]);
        }

        /**
         * 初始化是关键，只有j==0的时候全部赋为0，代表一开始的最大值为0，赋为-1，是为了
         * 保证最终选取m部片子，不多不少，如果中间少选了片子的话，是得到-1而已
         */
        for(int i=0;i<=l;i++){
            for(int j=0;j<=m;j++){

                dp[i][j]= (j==0 ? 0 : -1);
            }
        }

        /**
         * 在别人的代码中发现了新的优化技巧，那就是因为当j<c[i]的时候永远是跳过的，所以j的范围改成l~c[i]
         * 可以提高效率，也就是说将范围压缩在边界~当前物品的对应代价之间
         */
        for(int i=0;i<n;i++){
            for(int j=l;j>=c[i];j--){
                    for(int k=m;k>=1;k--){
                        if(dp[j-c[i]][k-1]>=0) dp[j][k]=max(dp[j][k] , dp[j-c[i]][k-1]+v[i]);
                        //printf("%d ",dp[j][k]);
                    }
                    //printf("    ");
            }
            //printf("
");
        }

        /**
         * 如果有最大值，一定会出现在dp[l][m]中，因为dp[l][]的的含义就是前l个物品中的最大值
         */
        if(dp[l][m]>=0) printf("%d
",dp[l][m]);
        else printf("0
");
    }
    return 0;
}

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