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HDU 1059 多重背包+二进制优化

2024-07-17 11:31:07   219人阅读

Dividing

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


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 0
1 0 0 0 1 1
0 0 0 0 0 0
 

 

Sample Output
Collection #1:
Can‘t be divided.
 
Collection #2:
Can be divided.
 
 
 
题目大意：
给你6个数分别为价值从1---6物品的个数，看能否把这些物品分为等价的两部分。
 
 
思路：
用多重背包，背包的体积和放的价值都是物品的价值。
单独多重背包很明显会超时，那么需要二进制优化了，所谓二进制优化就是把1种物品个数以二进制形式把该种物品弄成新的物品，在等价的条件下减少物品的个数，那么就可以用01背包写了，二进制优化看这个http://blog.csdn.net/weinierzui/article/details/23671419，写的很好。
 
代码：
 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 }

 


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