hdu 1712 ACboy needs your help(分組背包)

ACboy needs your help

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 4903    Accepted Submission(s): 2646

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
        

Source HDU 2007-Spring Programming Contest  

其實我覺得題目沒說一定隻能在一個地方選課複習= =

這題是分組背包。有時候題目中要求隻能在一堆物品裡面選一個，這樣的話就隻能給那些物品分組，然後在每一組裡面選一個。是以可以對于組别進行一次循環，然後每組裡面的物品都進行查找，比較最優者放入背包。

#include<stdio.h>
#include<string.h>
int max(int a,int b)
{
    if(a>b)return a;
    else return b;
}
int main()
{
    int n,m,i,j,k,A[105][105],dp[105];
    while(scanf("%d%d",&n,&m)!=EOF)
    {
    
        if(n==0&&m==0)break;
        memset(dp,0,sizeof(dp));
        for(i=1;i<=n;i++)
        for(j=1;j<=m;j++)
        scanf("%d",&A[i][j]);
        for(i=1;i<=n;i++)    //分組
        for(j=m;j>=1;j--)
        for(k=1;k<=m;k++)    //一組裡面每個物品都進行決策優化
        {
        if(j>=k)
        dp[j]=max(dp[j],dp[j-k]+A[i][k]);
    //    printf("%d\n",dp[j]);
        }
        printf("%d\n",dp[m]);
    }
}