HDOJ 1059 Dividing (多重背包二进制优化转01背包)

Dividing

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21606    Accepted Submission(s): 6094

Problem Description

Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbles. This would be easy if all the marbles had the same value, because then they could just split the collection in half. But unfortunately, some of the marbles are larger, or more beautiful than others. So, Marsha and Bill start by assigning a value, a natural number between one and six, to each marble. Now they want to divide the marbles so that each of them gets the same total value.
Unfortunately, they realize that it might be impossible to divide the marbles in this way (even if the total value of all marbles is even). For example, if there are one marble of value 1, one of value 3 and two of value 4, then they cannot be split into sets of equal value. So, they ask you to write a program that checks whether there is a fair partition of the marbles.

 

Input

Each line in the input describes one collection of marbles to be divided. The lines consist of six non-negative integers n1, n2, ..., n6, where ni is the number of marbles of value i. So, the example from above would be described by the input-line ``1 0 1 2 0 0''. The maximum total number of marbles will be 20000.

The last line of the input file will be ``0 0 0 0 0 0''; do not process this line.

 

Output

For each colletcion, output ``Collection #k:'', where k is the number of the test case, and then either ``Can be divided.'' or ``Can't be divided.''.

Output a blank line after each test case.

 

Sample Input

1 0 1 2 0 01 0 0 0 1 10 0 0 0 0 0

 

Sample Output

Collection #1:Can't be divided.Collection #2:Can be divided.

 


题意：给你第i种物品的数量，第i种物品价值为i，问能不能平分。。

思路：这道题用普通的多重转01的方法会超时，所以说不是正解，看到别人用二进制优化，就学习了一下。。

ac代码：

#include<stdio.h>#include<math.h>#include<string.h>#include<stack>#include<set>#include<queue>#include<vector>#include<iostream>#include<algorithm>#define MAXN 1010000#define LL long long#define ll __int64#define INF 0xfffffff#define mem(x) memset(x,0,sizeof(x))#define PI acos(-1)using namespace std;int a[7];int dp[MAXN];int num[MAXN];int main(){int i,j;int cas=0;while(scanf("%d%d%d%d%d%d",&a[1],&a[2],&a[3],&a[4],&a[5],&a[6])!=EOF){if(a[1]==a[2]&&a[2]==a[3]&&a[3]==a[4]&&a[4]==a[5]&&a[5]==a[6]&&a[6]==0)break;int sum=a[1]+a[2]*2+a[3]*3+a[4]*4+a[5]*5+a[6]*6;printf("Collection #%d:\n",++cas);if(sum%2){            printf("Can't be divided.\n\n");            continue;}int cnt=0;for(i=1;i<=6;i++){for(j=1;j<=a[i];j*=2){num[cnt++]=i*j;a[i]-=j;}if(a[i]>0)num[cnt++]=a[i]*i;}sum=sum/2;mem(dp);for(i=0;i<cnt;i++){for(j=sum;j>=num[i];j--)dp[j]=max(dp[j],dp[j-num[i]]+num[i]);}if(dp[sum]==sum)printf("Can be divided.\n\n");elseprintf("Can't be divided.\n\n");}return 0;}