分组背包问题描述与实现

#include <stdio.h>
int w[301][21];    

int get_max(int a, int b)
{
    if(a>=b)
        return a;
    else
        return b;
}

int main(void)
{
    int tc, T;
    
    
    setbuf(stdout, NULL);
    
    int Amout = 0;
    int company_num = 0;
    int i,j,k;
    
    scanf("%d", &T);
    for(tc = 0; tc < T; tc++)
    {
        
        scanf("%d%d", &Amout, &company_num);

        for(i = 1; i <= Amout; i++)
        {
            for(j = 0; j <= company_num; j++)
                scanf("%d", &w[i][j]);        //w[i][j] ：投入为i 时公司j能获得的利润。
        }

        int profit[21][301] = {0,};
        
        //分组背包问题        
        profit[1][1] = w[1][1];

        int max = profit[1][1];
        for( i = 2; i <= company_num; i++)    //amout = 1;
        {
            if (w[1][i] >max)    //w[k][i] ：投入为k时公司i能获得的利润
            {
                max = w[1][i];
                profit[i][1] = w[1][i];
            }
            else
                profit[i][1] = max;            
        }

        max = profit[1][1];
        for( i = 2; i <= Amout; i++)    //company = 1;
        {            
            if (w[i][1]  > max)
            {
                max = w[i][1];
                profit[1][i] = w[i][1];
            }
            else
            {
                profit[1][i] = max;
            }            
        }

        int t;
        //开始迭代运算了
        //f[k][v] = max{ f[k-1][v], f[k-1][v-c]+w }
        for( i = 2; i <= company_num; i++)
        {            
            for(j = 2; j <= Amout; j++)
            {                
                t =profit[i-1][j];                //不选第i个公司产品
                for( k =1; k<= j; k++)        //选i公司的产品
                {
                    if(profit[i-1][j-k] + w[k][i] > t)    //w[k][i] ：投入为k时公司i能获得的利润。
                        t = profit[i-1][j-k] + w[k][i];
                }
                profit[i][j] = t;                
            }
        }
        // Print the answer to standard output(screen).
        /*
        for( i =1; i<=company_num; i++)
            for(j =1; j<=Amout; j++)
            printf("profit[%d][%d] = %d\n",i ,j, profit[i][j]);
            */

        printf("%d\n", profit[company_num][Amout]);
    }

    return 0;//Your program should return 0 on normal termination.
}