hdu 1059 Dividing(多重背包)

 

http://acm.hdu.edu.cn/showproblem.php?pid=1059

 

解题思路：给你6种价值不一样的东西，并且每种物品都有一定的数量，现在问你，你可以把他们平均分成两份价值一样的吗？？

 

解题思路：多重背包问题

 

比较笨拙的代码

 

#include <stdio.h>#include <string.h>#include <math.h>#define Max(a,b) a>b?a:b#define Size 100005int c[2*Size],w[2*Size],dp[200005];int main(){int sum,i,num[7],no = 1,len,j,k,tem;while (1){sum = 0;for (i=1;i<=6;i++){scanf("%d",&num[i]);sum+=num[i]*i;}if(!sum)break;if(sum&1)else{len = 0;for (i=1;i<=6;i++){/*二进制划分,转成01背包做*/if(num[i]){k =(int)(log(num[i]+1)/log(2))-1;for (j=0;j<=k;j++){c[len] = w[len] = (int)(pow(2.0,j)*i);len++;}tem = (int)(pow(2.0,k+1))-1;if(num[i]!=tem){c[len] = w[len] = (num[i]-tem)*i;len++;}}}for(i=0;i<=sum;i++)dp[i] = 0;for (i=0;i<len;i++){for (j = sum/2;j>=c[i];j--){dp[j] = Max(dp[j],dp[j-c[i]]+w[i]);}}if(dp[sum/2] == sum/2)printf("Collection #%d:/nCan be divided./n/n",no++);elseprintf("Collection #%d:/nCan't be divided./n/n",no++);}}return 0;}

 

参考背包问题后，学习的代码

 

#include <stdio.h>#include <string.h>#define Size 60005#define Max(a,b) a>b?a:bint sum;int dp[Size];/*01背包*/void ZeroOnePack(int cost,int weight){int i;for(i = sum;i>=cost;i--)dp[i] = Max(dp[i],dp[i-cost]+weight);}/*完全背包*/void CompletePack(int cost,int weight){int i;for(i=cost;i<=sum;i++)dp[i] = Max(dp[i],dp[i-cost]+weight);}/*多重背包*/void MulPack(int cost,int weight,int count){int i;if(cost*count>=sum){CompletePack(cost,weight);return ;}i = 1;/*二进制划分*/while (i<count){ZeroOnePack(i*cost,i*weight);count-=i;i<<=1;}ZeroOnePack(count*cost,count*weight);}int main(){int num[7],no = 1,i;while (1){sum = 0;for(i=1;i<=6;i++){scanf("%d",&num[i]);sum+=num[i]*i;}if(!sum)break;if(sum&1)printf("Collection #%d:/nCan't be divided./n/n",no++);else{sum>>=1;for(i=0;i<=sum;i++)dp[i] = 0;for(i=1;i<=6;i++)if(num[i])MulPack(i,i,num[i]);if(dp[sum] == sum)printf("Collection #%d:/nCan be divided./n/n",no++);elseprintf("Collection #%d:/nCan't be divided./n/n",no++);}}return 0;}