HDU1059+多重背包

多重背包

题意：价值为1，2，3，4，5，6. 分别有n[1],n[2],n[3],n[4],n[5],n[6]个。

求能否找到满足价值刚好是所有的一半的方案。

/*
多重背包
*/
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<math.h>
using namespace std;
const int maxn = 10;
int n[ maxn ];
int sumv,V;
int dp[ 60011 ];
void bag01( int cost,int weight ){
    for( int i=V;i>=cost;i-- ){
        dp[ i ]=max(dp[ i ],dp[ i-cost ]+weight);
    }
    return ;
}
void bagall( int cost ,int weight ){
    for( int i=cost;i<=V;i++ ){
        dp[ i ]=max( dp[ i ],dp[ i-cost ]+weight);
    }
    return ;
}

void mul_bag( int cost,int weight,int num ){
    if( cost*num>=V ){
        bagall( cost,weight );
        return ;
    }
    int k=1;
    while( k<=num ){
        bag01( k*cost,k*weight );
        num-=k;
        k*=2;
    }
    bag01( num*cost,num*weight );
    return ;
}
int main()
{
    int i,k=1;
    while(cin>>n[0]>>n[1]>>n[2]>>n[3]>>n[4]>>n[5],n[0]+n[1]+n[2]+n[3]+n[4]+n[5])
    {
        memset(dp,0,sizeof(dp));
        sumv=n[0]+n[1]*2+n[2]*3+n[3]*4+n[4]*5+n[5]*6;
        if(sumv%2==1)
        {
            printf("Collection #%d:\nCan't be divided.\n\n",k++);
            continue;
        }
        V=sumv/2;
        for(i=0;i<6;i++)
        {
            if(n[i]) mul_bag(i+1,i+1,n[i]);
        }
        if(V==dp[V]) printf("Collection #%d:\nCan be divided.\n\n",k++);
        else printf("Collection #%d:\nCan't be divided.\n\n",k++);
    }
    return 0;
}