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): 5933    Accepted Submission(s): 3240

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

 

题意：ACBOY有n种功课要做，他总共有m天做这些功课,如果利用j天在功课i上，他能够获得的最大价值是a[i][j]问ACBOY如何才能获得最大价值.

分析：这题ACBOY有n种不同的功课，每种功课最多做一次，典型的分组背包问题.
        dp[i][j] 表示在前i门功课中花费j天能够获得的最大价值,当然写的时候压缩成一维.
解释一下分组背包的三层循环:
如果这样写： 那么前两层循环表示在第k组里面做01背包,第三层循环表示在第k组
里面选择某个物品,这样做的话能够保证在第 k 个组最多选一个。
for(k=1;k<=K;k++)
    for(int v = V;v>=0;v--)
        for(所有的i属于第k组)
如果这样写：后两层循环可以看成一个01背包,我们对每组里面的所有物品进行01背包
,这样就能够保证在第 k 组里面至少选一件.
for(k=1;k<=K;k++)
    for(所有的i属于第k组)
        for(int v = V;v>=0;v--)
至于其他的，碰到再说...

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#define N 105
using namespace std;

int dp[N];
int a[N][N];
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF,n+m){
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++) scanf("%d",&a[i][j]);
        }
        memset(dp,0,sizeof(dp));
        for(int k=1;k<=n;k++){
            for(int i=m;i>=0;i--){
                for(int j=1;j<=i;j++){   ///枚举第 k 组里面所有的天数
                    dp[i] = max(dp[i],dp[i-j]+a[k][j]);
                }
            }
        }
        printf("%d\n",dp[m]);
    }
    return 0;
}