HDU--1712--ACboy needs your help--分组背包/DP

ACboy needs your help

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

Total Submission(s): 4567    Accepted Submission(s): 2450

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 21 21 32 22 12 12 33 2 13 2 10 0

 

Sample Output

346

 

题意：有n堂课，每堂课可以用不懂的天数完成，然后给你一个数组表d[x][y]表示第x堂课用y天来完成可以得到的价值，一共给你m天，每堂课只能完成一次，求能得到的最大价值。

题解：这是这两天开始做题以后第一个让我想博客一下的，这题用的是分组背包，看了下九讲里面的，就写了，然后WA，于是琢磨了半天才发现一个细节，分组背包的要点就是遍历组内物品的循环一定要放在选取背包容量的循环里面。

#include <iostream>#include <cstdio>#include <cstring>#define Max(a,b) a>b?a:busing namespace std;int dp[11111];int main (void){    int n,m,i,j,k,l,d[111][111];    while(scanf("%d%d",&n,&m)&&n+m>0)    {        memset(dp,0,sizeof(dp));        for(i=1;i<=n;i++)        for(j=1;j<=m;j++)        scanf("%d",&d[i][j]);        for(i=1;i<=n;i++)        {            for(k=m;k>=1;k--)            for(j=1;j<=k;j++)            {                dp[k]=Max(dp[k],dp[k-j]+d[i][j]);            }        }        printf("%d\n",dp[m]);    }    return 0;}

总结：OLE了两次，我把用来显示每个物品运行完之后dp数组内数值的那句忘记删掉了。。。