ACboy needs your help (HDU_1712) 分组背包

ACboy needs your help

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5807    Accepted Submission(s): 3171

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

题目大意：共有m天时间，已知有n门课程，每门课程学习天数从1-m都有着不同的价值，求在这m天内做到能获得多少价值？

解题思路：分组背包。在原来01背包的基础上再加上一层循环。

#include"cstdio"#include"cstring"using namespace std;const int maxn=105;int dp[maxn];int value[maxn];int cost[maxn];int m;int max(int x,int y){return x>y?x:y;}int main(){int n;while(1){scanf("%d%d",&n,&m);if(n==0&&m==0) break;memset(dp,0,sizeof(dp));for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){scanf("%d",&value[j]);cost[j]=j;}for(int j=m;j>=1;j--){for(int k=1;k<=m;k++){if(j>=cost[k]){dp[j]=max(dp[j],dp[j-cost[k]]+value[k]);}}}}printf("%d\n",dp[m]);}return 0;}