HDOJ1059(多重背包)

1.解法一：多重背包

#include<iostream>
#include<cstdio>
using namespace std;
#define MAX(a,b) (a>b)?a:b
const int SIZE=+;
int dp[SIZE];
int bag[];
int nLimit;

void ZeroOnePack(int cost, int value)
{
    for(int i=nLimit; i>=cost; i--)
        dp[i]=MAX(dp[i],dp[i-cost]+value);
}

void CompletePack(int cost, int value)
{
    for(int i=cost; i<=nLimit; i++)
        dp[i]=MAX(dp[i],dp[i-cost]+value);
}

void MultiplyPack(int cost, int value, int amount)
{
    if(amount*cost>=nLimit) CompletePack(cost,value);
    else
    {
        int k=;
        while(k<amount)
        {
            ZeroOnePack(k*cost, k*value);
            amount-=k;
            k<<=;
        }

        ZeroOnePack(amount*cost,amount*value);
    }
}

bool CheckFinish()
{
    for(int i=;i<=;i++)
        if(bag[i]!=)
            return false;
    return true;
}

int main()
{
    int T=;
    while(true)
    {
        int sum=;
        for(int i=; i<=; i++)
        {
            scanf("%d",&bag[i]);
            sum+=bag[i]*i;
        }
        if(CheckFinish())
            break;

        printf("Collection #%d:\n",++T);
        if(sum%==)
        {
            printf("Can't be divided.\n");
        }
        else
        {
            memset(dp,,sizeof(dp));
            nLimit=sum/;
            for(int i=;i<=;i++)
            {
                MultiplyPack(i,i,bag[i]);
            }
            if(dp[nLimit]==nLimit)
                printf("Can be divided.\n");
            else
                printf("Can't be divided.\n");
        }
        printf("\n");

    }

    return ;
}

2.解法二：多重部分和

#include<cstdio>
#include<cstring>
using namespace std;
const int SIZE=+;
int a[];
int dp[SIZE];
bool check()
{
    for(int i=;i<;i++)
        if(a[i]!=)
            return true;
    return false;
}
int sum;

int main()
{
    int t=;
    while(true)
    {
        sum=;
        for(int i=;i<;i++)
        {
            scanf("%d",&a[i]);
            sum+=(i+)*a[i];
        }
        if(!check())
            break;    

        printf("Collection #%d:\n",++t);
        if(sum%==)
        {
            printf("Can't be divided.\n");
        }
        else
        {
            memset(dp,-,sizeof(dp));
            int k=sum/;
            dp[]=;
            for(int i=;i<;i++)
            {
                for(int j=;j<=k;j++)
                {
                    if(dp[j]>=)
                    {
                        dp[j]=a[i];
                    }
                    else if(j<(i+)||dp[j-(i+)]<=)
                    {
                        dp[j]=-;
                    }
                    else
                    {
                        dp[j]=dp[j-(i+)]-;
                    }
                }
            }
            if(dp[k]>=)
            {
                printf("Can be divided.\n");
            }
            else
            {
                printf("Can't be divided.\n");
            }
        }
        printf("\n");
    }
    return ;
}

3.解法三：判断多重背包可否装满

#include<iostream>
#include<cstdio>
using namespace std;
#define MAX(a,b) (a>b)?a:b
const int SIZE=+;
const int INF=;
int dp[SIZE];
int bag[];
int nLimit;
bool CheckFinish()
{
    for(int i=;i<=;i++)
        if(bag[i]!=)
            return false;
    return true;
}

int main()
{
    int T=;
    while(true)
    {
        int sum=;
        for(int i=; i<=; i++)
        {
            scanf("%d",&bag[i]);
            sum+=bag[i]*i;
        }
        if(CheckFinish())
            break;

        printf("Collection #%d:\n",++T);
        if(sum%==)
        {
            printf("Can't be divided.\n");
        }
        else
        {
            nLimit=sum/;
            memset(dp,,sizeof(dp));
            dp[]=;
            int dpt[];
            for(int i=;i<=;i++)
            {
                memset(dpt,,sizeof(dpt));
                for(int j=i;j<=nLimit;j++)
                    if(!dp[j]&&dp[j-i]&&dpt[j-i]<bag[i])
                    {
                        dpt[j]=dpt[j-i]+;
                        dp[j]=;
                    }
            }

            if(dp[nLimit])
            {
                printf("Can be divided.\n");
            }
            else
            {
                printf("Can't be divided.\n");
            }
        }

        printf("\n");
    }

    return ;
}

判断模板：

memset(dp,,sizeof(dp));
dp[]=;
int used[];
for(int i=;i<=nKind;i++)
{
    memset(used,,sizeof(used));
    for(int j=weight[i];j<=nLimit;j++)
        if(!dp[j]&&dp[j-weight[i]]&&used[j-weight[i]]<bag[i])
        {
            used[j]=used[j-weight[i]]+;
            dp[j]=;
        }
}