HDOJ 1059 Dividing

最新推荐文章于 2019-03-07 21:15:54 发布

转载最新推荐文章于 2019-03-07 21:15:54 发布 · 149 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/CKboss/archive/2013/06/06/3351036.html

文章标签：

#java

本文探讨了一种特殊的背包问题，即如何将不同价值的宝石公平地分配给两个人。通过使用多重背包问题的方法，实现了一个高效的算法来解决这个问题。文章提供了一个具体的例子，并附带了完整的代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

惊现多重背包！马上水掉

Dividing

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12116 Accepted Submission(s): 3394

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

Sample Output

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

Source

Mid-Central European Regional Contest 1999

Recommend

JGShining

#include <iostream>

#include <cstring>

#include <cstdio>

using namespace std;

int a[7];

int pack[422222];

int V;

void zpack(int*a,int c,int w)

{

for(int i=V;i>=c;i--)

a =max(a,a[i-c]+w);

}

void cpack(int*a,int c,int w)

{

for(int i=c;i<=V;i++)

a =max(a,a[i-c]+w);

}

void multipack(int* a,int c,int w,int m)

{

if(c*m>=V)

{

cpack(a,c,w);

return;

}

int k=1;

while(k<m)

{

zpack(a,k*c,k*w);

m-=k;

k*=2;

}

zpack(a,m*c,m*w);

}

int main()

{

a[0]=1;

int cur=0;

while(a[0])

{

int sum=0;

for(int i=1;i<=6;i++)

{

scanf("%d",&a );

sum+=i*a ;

}

if(sum==0) a[0]=0;

if(a[0]==0) break;

// cout<<sum<<endl;

if(sum&1) { printf("Collection #%d:\n",++cur); puts("Can't be divided.\n"); continue;}

memset(pack,0,sizeof(pack));

V=sum/2;

for(int i=1;i<=6;i++)

multipack(pack,i,i,a );

for(int i=0;i<=V;i++)

cout<<pack <<" ";

cout<<endl;

if(pack[V]==V)

{

printf("Collection #%d:\n",++cur); puts("Can be divided.\n");

}

else

{

printf("Collection #%d:\n",++cur); puts("Can't be divided.\n");

}

return 0;

}

转载于:https://www.cnblogs.com/CKboss/archive/2013/06/06/3351036.html