hdu 1712 ACboy needs your help

標籤：hdu   分組背包   

轉載請註明出處：http://blog.csdn.net/u012860063

題目連結：http://acm.hdu.edu.cn/showproblem.php?pid=1712

Problem DescriptionACboy 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?
 InputThe 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.
 OutputFor 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

題意：一開始輸入n和m，n代表有n門課，m代表你有m天，
然後給你一個數組，val[i][j]，代表第i門課，在通過j天去修，
會得到的分數。求在m天能得到的最大分數。

#include <cstdio>#include <cstring>#define N 147int max(int a, int b){if(a > b)return a;return b;}int main(){int n,m,a[N][N],dp[N];int i, j, k;while(~scanf("%d%d",&n,&m)&&( n || m)){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++)//第一重迴圈：分組數{for(j = m ; j >= 1 ; j--) //第二重迴圈：容量體積{for(k = 1 ; k <= j ; k++) //第三重迴圈：屬於i組的k{dp[j]=max(dp[j],dp[j-k]+a[i][k]);}}}printf("%d\n",dp[m]);}return 0;}