B - ACboy needs your help（动态规划，分组背包）

B - ACboy needs your help

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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Description

ACboy has N courses this term, and he plans to spend at most M days on study.Of course,the profit he will gain from different course depending on the days he spend on it.How to arrange the M days for the N courses to maximize the profit?

 

Input

The input consists of multiple data sets. A data set starts with a line containing two positive integers N and M, N is the number of courses, M is the days ACboy has.
Next follow a matrix A[i][j], (1<=i<=N<=100,1<=j<=M<=100).A[i][j] indicates if ACboy spend j days on ith course he will get profit of value A[i][j].
N = 0 and M = 0 ends the input.

 

Output

For each data set, your program should output a line which contains the number of the max profit ACboy will gain.

 

Sample Input

2 2 1 2 1 3 2 2 2 1 2 1 2 3 3 2 1 3 2 1 0 0

 

Sample Output

3

4

6

题意:

ACboy有n门课可以参加，他有m天时间。一天只能参加一门课。一门课可以参加多天，参加的天数不同获得的经验不一样，并且一门课只能修习一次。但是要注意的是，同一门课不是参加的天数越多，经验就越高。输入的表示是，行数表示第几门课，列数表示上几天，该数就是第几门课参加几天获得的经验数。问m天能获得最多的经验数。

思路：分组背包，建立dp数组dp[i][j]表示修习前i种课程，修习了j天，所得的最大的经验值。

二维数组：

方法：状态转移方程为dp[i][j]=max{dp[i-1][j],dp[i-1][j-k]+xing[i][k]},意思是前i-1种课程，前j-k天获得的经验最大数是dp[i-1][j-k]，后k天参加第i门课获得的经验是xing[i][k]。比较此时的dp[i][j]和dp[i-1][j-k]+xing[i][k]谁大。

AC代码：

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstdlib>
 4 #include<cstring>
 5 
 6 using namespace std;
 7 
 8 
 9 int dp[103][103]={0};
10 int xing[103][103]={0};
11 
12 
13 int main()
14 {
15 //    freopen("1.txt","r",stdin);
16     int n,m;
17     while(cin>>n>>m&&n!=0&&m!=0){
18         memset(dp,0,sizeof(dp));
19         int i,j,k;
20         for(i=1;i<=n;i++){//输入数据
21             for(j=1;j<=m;j++)
22                 cin>>xing[i][j];
23         }
24         for(i=1;i<=n;i++){
25             for(j=1;j<=m;j++)
26             for(k=0;k<=j;k++){
27                 if(dp[i][j]<=dp[i-1][j-k]+xing[i][k])//进行比较
28                     dp[i][j]=dp[i-1][j-k]+xing[i][k];//状态转移
29             }
30         }
31         cout<<dp[n][m]<<endl;
32     }
33     return 0;
34 }

View Code

一位数组：

方法：状态转移方程为dp[j]=max{dp[[j],dp[j-k]+xing[i][k]},意思是前j-k天获得的经验最大数是dp[j-k]，后k天参加第i门课获得的经验是xing[i][k]。比较此时的dp[j]和dp[j-k]+xing[i][k]谁大。

AC代码：

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstdlib>
 4 #include<cstring>
 5 
 6 using namespace std;
 7 
 8 
 9 int dp[103]={0};
10 int xing[103][103]={0};
11 
12 
13 int main()
14 {
15 //    freopen("1.txt","r",stdin);
16     int n,m;
17     while(cin>>n>>m&&n!=0&&m!=0){
18         memset(dp,0,sizeof(dp));
19         int i,j,k;
20         for(i=1;i<=n;i++){
21             for(j=1;j<=m;j++)
22                 cin>>xing[i][j];
23         }
24         for(i=1;i<=n;i++){
25             for(j=m;j>=1;j--)
26             for(k=0;k<=j;k++){
27                 if(dp[j]<=dp[j-k]+xing[i][k])
28                     dp[j]=dp[j-k]+xing[i][k];
29             }
30         }
31         cout<<dp[m]<<endl;
32     }
33     return 0;
34 }

View Code

 

转载于:https://www.cnblogs.com/zhangchengbing/p/3353409.html