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): 4912    Accepted Submission(s): 2651

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

 

 

题目意思：
有n个课程，每个课程在不同天数得到不同价值，总共m天，问m天之内得到最大的价值。

 

思路：
每个课程若在i天完成，就不能在j天完成。以n个课程为组，天数为体积，价值为价值，即得到分组背包的模型。

 

dp[i][j]=max(dp[i-1][j],dp[i-1][j-v[k]]+w[k])//dp[i][j]表示前i组已占用j体积时最大价值。

 

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <queue>
#include <cmath>
#include <set>
using namespace std;

#define N 105

int max(int x,int y){return x>y?x:y;}
int min(int x,int y){return x<y?x:y;}
int abs(int x,int y){return x<0?-x:x;}


int a[N][N];
int n, m;
int dp[N];

main()
{
    int i, j, k;
    while(scanf("%d %d",&n,&m)==2){
        if(!n&&!m) break;
        for(i=1;i<=n;i++){
            for(j=1;j<=m;j++)
                scanf("%d",&a[i][j]);
        }
        memset(dp,0,sizeof(dp));
        for(i=1;i<=n;i++){
            for(j=m;j>=0;j--){
                for(k=1;k<=j;k++){
                    dp[j]=max(dp[j],dp[j-k]+a[i][k]);
                }
            }
        }
        printf("%d\n",dp[m]);
    }
}