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 ;
}