HDOJ 1171 Big Event in HDU(多重背包转化01背包 or 母函数)

Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 29792    Accepted Submission(s): 10451

Problem Description

Nowadays, we all know that Computer College is the biggest department in HDU. But, maybe you don't know that Computer College had ever been split into Computer College and Software College in 2002.
The splitting is absolutely a big event in HDU! At the same time, it is a trouble thing too. All facilities must go halves. First, all facilities are assessed, and two facilities are thought to be same if they have the same value. It is assumed that there is N (0<N<1000) kinds of facilities (different value, different kinds).

 

Input

Input contains multiple test cases. Each test case starts with a number N (0 < N <= 50 -- the total number of different facilities). The next N lines contain an integer V (0<V<=50 --value of facility) and an integer M (0<M<=100 --corresponding number of the facilities) each. You can assume that all V are different.
A test case starting with a negative integer terminates input and this test case is not to be processed.

 

Output

For each case, print one line containing two integers A and B which denote the value of Computer College and Software College will get respectively. A and B should be as equal as possible. At the same time, you should guarantee that A is not less than B.

 

Sample Input

2
10 1
20 1
3
10 1 
20 2
30 1
-1

 

Sample Output

20 10
40 40

 

多重背包就是当做01背包求解，代码如下：(注意题目是输入负数结束，不是-1)

<span style="font-size:18px;">#include<cstdio>
#include<cstring>
int dp[130000],val[5010];
int main()
{
	int n,m,sum,cnt,i,j,k;
	while(scanf("%d",&n)!=EOF)
	{
		if(n<0)
		   break;
		k=0;sum=0;
		while(n--)
		{
			scanf("%d%d",&val[k],&m);
			sum+=val[k]*m;
			m--;
			k++;
			while(m--)
			{
				val[k]=val[k-1];
				k++;
			}
		}
		cnt=sum;
		cnt>>=1;
	    memset(dp,0,sizeof(dp));
	    for(i=0;i<k;++i)
	    {
		    for(j=cnt;j>=val[i];j--)
		       dp[j]=dp[j]>dp[j-val[i]]+val[i]?dp[j]:dp[j-val[i]]+val[i];
	    }
	    printf("%d %d\n",sum-dp[cnt],dp[cnt]);
	}
	return 0;
} </span>

母函数解法： 刚学的母函数也能解，找到指数小于sum/2的系数不为0的最大指数，即分配中比较小的部分。

代码如下:

#include<cstdio>
#include<cstring>
#define maxn 2500100
int val[55],num[55],c1[maxn],c2[maxn];
int main()
{
	int i,j,k,n,sum,used;
	while(scanf("%d",&n)&&n>=0)
	{
		sum=0;
		for(i=1;i<=n;++i)//注意表达式的序号从1开始计数 
		{
			scanf("%d%d",&val[i],&num[i]);
			sum+=val[i]*num[i];
		}
		memset(c1,0,sizeof(c1));
		memset(c2,0,sizeof(c2));
		for(i=0,j=0;j<=num[1];j++,i+=val[1])
			c1[i]=1;
		for(i=2;i<=n;++i)
		{
			for(j=0;j<=sum/2;++j)
			{
				for(k=0,used=0;k+j<=sum/2&&used<=num[i];used++,k+=val[i])
					c2[k+j]+=c1[j];
			}
			for(j=0;j<=sum/2;++j)
			{
				c1[j]=c2[j];  c2[j]=0;
			}
		}
		for(i=sum/2;i>=0;i--)
		{
			if(c1[i])
			{
				printf("%d %d\n",sum-i,i);
				break;
			}
		}
	}
	return 0;
}