HDU 1059 多重背包+二進位最佳化

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Dividing

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16909    Accepted Submission(s): 4729

Problem DescriptionMarsha 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. 

 

InputEach 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. 

 

OutputFor 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 Input1 0 1 2 0 01 0 0 0 1 10 0 0 0 0 0 

 

Sample OutputCollection #1:Can‘t be divided. Collection #2:Can be divided.   題目大意：給你6個數分別為價值從1---6物品的個數，看能否把這些物品分為等價的兩部分。  思路：用多重背包，背包的體積和放的價值都是物品的價值。單獨多重背包很明顯會逾時，那麼需要二進位最佳化了，所謂二進位最佳化就是把1種物品個數以二進位形式把該種物品弄成新的物品，在等價的條件下減少物品的個數，那麼就可以用01背包寫了，二進位最佳化看這個http://blog.csdn.net/weinierzui/article/details/23671419，寫的很好。 代碼：

 1 #include <cstdio> 2 #include <cstring> 3 #include <vector> 4 #include <algorithm> 5 #include <iostream> 6 using namespace std; 7  8 int dp[100000]; 9 int v[30];10 11 main()12 {13     int n, i, j, k, sum, kase=1;14     int a[7];15     while(scanf("%d %d %d %d %d %d",&a[1],&a[2],&a[3],&a[4],&a[5],&a[6])==6&&(a[1]+a[2]+a[3]+a[4]+a[5]+a[6])){16         n=0;17     memset(dp,0,sizeof(dp));18         printf("Collection #%d:\n",kase++);19         sum=a[1]+a[2]*2+a[3]*3+a[4]*4+a[5]*5+a[6]*6;20         if(sum&1){21             printf("Can‘t be divided.\n\n");continue;22         }23         for(i=1;i<=6;i++){24             n=0;25             for(j=1;j<=a[i];j<<=1){26                 v[n++]=j;27                 a[i]-=j;28             }29             if(a[i]>0){30                 v[n++]=a[i];31             }32             for(j=0;j<n;j++){33             for(k=sum/2;k>=v[j]*i;k--)34             dp[k]=max(dp[k],dp[k-v[j]*i]+v[j]*i);35           }36         }37         38         if(dp[sum/2]==sum/2){39             printf("Can be divided.\n\n");40         }41         else printf("Can‘t be divided.\n\n");42     }43 }