HDU

先上题目：

ACboy needs your help

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3464    Accepted Submission(s): 1792

Problem 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：主角有n门课，m天，不同的课花不同的时间在里面会有不同的价值，问最大价值是多少？

　　分组背包，状态转移方程dp[k][v]=max{dp[k-1][v],dp[k-1][v-c[k][i]]+w[k][i]}    c[k][i],w[k][i]分别代表第k组物品的i个物品的代价和价值。

　　需要注意的问题是，枚举背包容量和每一个分组里面的物品的时候，最外面一层循环枚举组数，中间一层应该是枚举背包容量，里面一层枚举某一个分组里面的物品，为什么要这样，因为如果中间一层和最里面的一层如果交换了位置的话，那就不是01背包了，同一组里面的物品被多次枚举，而原本同一组里面的物品需要互斥，所以最里面一层枚举某一个分组里面的物品，这样的话因为枚举背包容量的时候是从大到小枚举的，所以对于同一个背包容量，某一组的物品只会放进去一次，这样就保证互斥了。

 

上代码：

 

#include <cstdio>
#include <cstring>
#define max(x,y) (x > y ? x : y)
#define MAX 102
using namespace std;

int a[MAX][MAX];
int dp[MAX];

int main()
{
    int n,m,maxn;
    //freopen("data.txt","r",stdin);
    while(scanf("%d %d",&n,&m),(n+m)){
        memset(a,0,sizeof(a));
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                scanf("%d",&a[i][j]);
            }
        }
        maxn=0;
        for(int i=1;i<=n;i++){
            for(int k=m;k>=1;k--){
                for(int j=1;j<=k;j++){
                    dp[k]=max(dp[k],dp[k-j]+a[i][j]);
                    maxn=max(dp[k],maxn);
                }
            }
        }
        printf("%d
",maxn);
    }
    return 0;
}