HDU1712-ACboy needs your help

描述：

　　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?

　　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.

　　For each data set, your program should output a line which contains the number of the max profit ACboy will gain.

代码：

　　多重背包问题，不要求装满背包。

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<stdlib.h>
#include <math.h>
using namespace std;
#define MAX 105

int main(){
    int n,m,dp[MAX][MAX],value[MAX][MAX],max;
    while( scanf("%d%d",&n,&m)!=EOF && n!= && m!= ){
        memset(value,,sizeof(value));
        for( int i=;i<=n;i++ )
            for( int j=;j<=m;j++ )
                scanf("%d",&value[i][j]);
        memset(dp,,sizeof(dp));//并未要求背包放满

        for( int i=;i<=n;i++ ){
            for( int j=;j<=m;j++ ){
                max=;
                for( int k=;k<=j;k++ )//第i个物品放0-j个
                    max=(max>dp[i-][j-k]+value[i][k])?max:dp[i-][j-k]+value[i][k];
                dp[i][j]=max;//背包容量为j，放前i个物品得到的最大值
            }
        }
        printf("%d\n",dp[n][m]);
    }
    system("pause");
    return ;
}

　　空间复杂度优化。可以看出，当计算dp[i][j]时，用到的数值为dp[i-1][0]到dp[i-1][j]的值，故背包容量的遍历顺序需反序，可以将dp二维数组优化为一维。

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<stdlib.h>
#include <math.h>
using namespace std;
#define MAX 105

int main(){
    int n,m,dp[MAX],value[MAX][MAX],max;
    while( scanf("%d%d",&n,&m)!=EOF && n!= && m!= ){
        memset(value,,sizeof(value));
        for( int i=;i<=n;i++ )
            for( int j=;j<=m;j++ )
                scanf("%d",&value[i][j]);
        memset(dp,,sizeof(dp));//并未要求背包放满

        for( int i=;i<=n;i++ ){
            for( int j=m;j>=;j-- ){
                max=;
                for( int k=;k<=j;k++ )//第i个物品放0-j个
                    max=(max>dp[j-k]+value[i][k])?max:dp[j-k]+value[i][k];
                dp[j]=max;//背包容量为j，放前i个物品得到的最大值
            }
        }
        printf("%d\n",dp[m]);
    }
    system("pause");
    return ;
}