HDU——1059Dividing（母函数或多重背包）

最新推荐文章于 2024-07-28 10:45:00 发布

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最新推荐文章于 2024-07-28 10:45:00 发布

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文章标签： java

原文链接：http://www.cnblogs.com/Blackops/p/5766303.html

本文介绍了一个公平分配不同价值大理石的问题，通过两种不同的编程方法解决这一挑战：一是使用母函数的方法来检查是否可以将大理石分为两个等值集合；二是采用多重背包问题的解法，通过动态规划实现解决方案。

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Dividing

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23277 Accepted Submission(s): 6616

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

0 1 2 0 0
0 0 0 1 1
0 0 0 0 0

Sample Output

Collection #1:
Can't be divided.

Collection #2:
Can be divided.

看看能否凑出V/2这个值就可以了。母函数比较好理解

代码：

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define MM(x) memset(x,0,sizeof(x))
#define MMINF(x) memset(x,INF,sizeof(x))
typedef long long LL;
const double PI=acos(-1.0);
const int N=1300;
int c[7];
int c1[N],c2[N];
int main(void)
{
	int i,j,k,V,tcase=0;
	while (~scanf("%d%d%d%d%d%d",&c[1],&c[2],&c[3],&c[4],&c[5],&c[6]))
	{
		V=0;
		for (i=1; i<=6; i++)
		{
			c[i]%=60;
			V+=c[i]*i;
		}
		if(!V)
			break;
		if(V&1)
		{
			printf("Collection #%d:\nCan't be divided.\n\n",++tcase);
			continue;
		}
		V>>=1;
		MM(c1);MM(c2);
		int index=0;
		for (i=1; i<=6; i++)
		{
			if(c[i])
			{
				for (j=0; j<=c[i]; j++)
				{
					c1[j*i]=1;
					index=i;
				}
				break;
			}
		}
		for (i=index+1; i<=6; i++)
		{
			for (j=0; j<=V; j++)
			{
				if(c1[j])
				{
					for (k=0; k<=c[i]; k++)
					{
						if(j+i*k>V)
							break;
						c2[j+i*k]+=c1[j];
					}
				}	
			}
			memcpy(c1,c2,sizeof(c2));
			MM(c2);
		}
		printf("Collection #%d:\n",++tcase);
		puts(c1[V]?"Can be divided.\n":"Can't be divided.\n");
		MM(c);
	}
	return 0;
}

然后是多重背包的。感觉以后这样的题还是写成自定义函数吧。不然课件上非常难看懂。

代码：

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define MM(x) memset(x,0,sizeof(x))
#define MMINF(x) memset(x,INF,sizeof(x))
typedef long long LL;
const double PI=acos(-1.0);
const int N=122000;
int dp[N];
int V;
int c[7];
void zonepack(int cost,int val)
{
	for (int i=V; i>=cost ; --i)
	{
		dp[i]=max(dp[i],dp[i-cost]+val);
	}
}
void wanquanpack(int cost,int val)
{
	for (int i=cost; i<=V ; ++i)
	{
		dp[i]=max(dp[i],dp[i-cost]+val);
	}
}
int main(void)
{
	int n,i,j,k,tcase=0,cnt;
	while (~scanf("%d%d%d%d%d%d",&c[1],&c[2],&c[3],&c[4],&c[5],&c[6])&&(c[1]||c[2]||c[3]||c[4]||c[5]||c[6]))
	{
		cnt=0;
		V=0;
		for (i=1; i<=6; i++)
			V+=c[i]*i;
		if(V&1)
		{
			printf("Collection #%d:\n%s\n\n",++tcase,"Can't be divided.");
			continue;
		}
		V>>=1;
		MM(dp);
		for (i=1; i<=6; i++)
		{
			if(c[i]*i>=V)
			{
				wanquanpack(i,i);
			}
			else
			{
				int k=1,t=c[i];
				while (k<t)
				{
					zonepack(k*i,k*i);
					t-=k;
					k<<=1;
				}
				zonepack(t*i,t*i);
			}
		}
		printf("Collection #%d:\n%s\n\n",++tcase,dp[V]!=V?"Can't be divided.":"Can be divided.");
	}
	return 0;
}

转载于:https://www.cnblogs.com/Blackops/p/5766303.html