POJ 1014 Dividing （多重背包）

题目链接：.........

题目大意：有6种价值的大理石（为1~6）， 先给出每种价值的大理石若干， 求能不能将它们平分。

思路：背包9讲中的模板套着打就ok了， dp[j]表示重量为j时的价值， 这里重量等于价值。 

            这题和上一道题很像（.........）， 也可以用f[j]记录重量为j时是否出现过。

code：

#include <stdio.h>
#include <string.h>
int sum = 0, c[12], dp[20002*6];
void zero_one(int weight, int value)
{
    int j = 0;
    for(j = sum; j>=weight; j--)
        dp[j] = dp[j]>dp[j-weight]+value? dp[j]:dp[j-weight]+value;
}
int main()
{
    int i = 0, j = 0, k = 0, count = 0;
    while(1)
    {
        sum = 0;
        for(i = 1; i<=6; i++)
        {
            scanf("%d", &c[i]);
            sum += i*c[i];
        }
        if(!sum)
            break;
        printf("Collection #%d:\n",++count);
        if(sum%2 == 1)
            printf("Can't be divided.\n");
        else
        {
            sum /= 2;
            memset(dp, 0, sizeof(dp));
            for(i = 1; i<=6; i++)
            {
                if(i*c[i]>sum)//完全背包
                {
                    for(j = i; j<=sum; j++)
                        dp[j] = dp[j]>dp[j-i]+i? dp[j]:dp[j-i]+i;
                }
                else
                {
                    k = 1;
                    while(k<c[i])//2进制拆分
                    {
                        zero_one(k*i, k*i);
                        c[i] -= k;
                        k *= 2;
                    }
                    zero_one(c[i]*i, c[i]*i);
                }
                if(dp[sum] == sum)
                    break;
            }
            if(dp[sum] == sum)
                printf("Can be divided.\n");
            else
                printf("Can't be divided.\n");
        }
        printf("\n");
    }
    return 0;
}

code：

#include <stdio.h>
#include <string.h>
int sum = 0, c[12], used[20002*6], f[20002*6];
int main()
{
    int i = 0, j = 0, k = 0, count = 0;
    while(1)
    {
        sum = 0;
        for(i = 1; i<=6; i++)
        {
            scanf("%d", &c[i]);
            sum += i*c[i];
        }
        if(!sum)
            break;
        printf("Collection #%d:\n",++count);
        if(sum%2 == 1)
            printf("Can't be divided.\n");
        else
        {
            sum /= 2;
            memset(f, 0, sizeof(f));
            f[0] = 1;
            for(i = 1; i<=6; i++)
            {
                memset(used, 0, sizeof(used));
                for(j = i; j<=sum; j++)
                {
                    if(!f[j] && f[j-i] && used[j-i]<c[i])
                    {
                        f[j] = 1;
                        used[j] = used[j-i]+1;
                    }
                }
                if(f[sum])
                    break;
            }
            if(f[sum])
                printf("Can be divided.\n");
            else
                printf("Can't be divided.\n");
        }
        printf("\n");
    }
    return 0;
}