hdu1712（01背包）

ACboy needs your help

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6054    Accepted Submission(s): 3309

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：ACboy要花m天来复习n门课，现在告诉你他花j天在第i门课上会得到的奖励，让你给他算出最优解。

思路：01背包稍微变形一下，其实是一模一样的。

dp[i]表示花i天来复习的最优解，然后转移方程是dp[j]=max(dp[j],dp[j-k]+val[i][k]);

其实想出状态转移方程不难，我们要做的就是一一对应01背包的各种条件，显然这题m天是背包容量，花k天来复习是消耗的容量，val[i][j]则是对应的得到的价值。

#include <iostream>#include <stdio.h>#include <stdlib.h>#include<string.h>#include<algorithm>#include<math.h>#include<queue>#include<stack>using namespace std;typedef long long ll;int val[110][110],dp[110];int main(){    int m,n;    while(~scanf("%d%d",&n,&m)&&m+n)    {        for(int i=1; i<=n; i++)            for(int j=1; j<=m; j++)                scanf("%d",&val[i][j]);        memset(dp,0,sizeof(dp));        for(int i=1; i<=n; i++)            for(int j=m; j>=0; j--)                for(int k=1; k<=j; k++)                    dp[j]=max(dp[j],dp[j-k]+val[i][k]);            printf("%d\n",dp[m]);    }    return 0;}