(DP,二维背包)Watch The Movie--HDOJ

Watch The Movie
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 893 Accepted Submission(s): 335

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
2010 ACM-ICPC Multi-University Training Contest（6）——Host by BIT

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zhouzeyong

总结：
之前没有遇到过这种类型的背包，想了一会后，便搜了题解。这是一道二维背包的题目，也算是一个模板吧，状态转移方程为
dp[j][k] = max(dp[j][k],dp[j-1][k-t[i]] + v[i]); j代表所包含的物品数量，k则代表所花费的时间，dp[j][k] 则是我们要求的最大价值
在初始化的时候，要把物品数为零的，即dp[0][] 初始化成0，其余的初始化成无穷小

#include<iostream>#include<stdio.h>#include<algorithm>#include<string.h>#include<math.h>using namespace std;int main(void){//    freopen("in.txt","r",stdin);    int ncase;    cin >>ncase;    while(ncase--)    {        int n,m,l;        scanf("%d %d %d",&n,&m,&l);        int v[n+1],t[n+1];        for(int i=0; i<n; ++i)        {            scanf("%d %d",&t[i],&v[i]);        }        int dp[n+1][l+1];        memset(dp,0,sizeof(dp));        for(int i=1; i<=n; ++i)        for(int j=0; j<=l; ++j)        {            dp[i][j] = -999999;        }        for(int i=0; i<n; ++i)        for(int j=m; j>=1; --j)        for(int k=l; k>=t[i]; --k)        {            dp[j][k] = max(dp[j][k],dp[j-1][k-t[i]] + v[i]);        }        if(dp[m][l] < 0)            cout << "0"<<endl;        else            cout <<dp[m][l]<<endl;    }    return 0;}