Big Event in HDU

最新推荐文章于 2020-09-04 01:51:25 发布

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最新推荐文章于 2020-09-04 01:51:25 发布

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原文链接：http://www.cnblogs.com/yskyskyer123/p/4513176.html

本文探讨了多重背包问题的解决方法，包括提供代码实现和母函数方法，并通过实例展示了如何分配不同设施的价值给计算机学院和软件学院，确保分配尽可能公平。

提供多重背包的一些思路

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

10 1

20 1

10 1

20 2

30 1

-1

Sample Output

20 10

40 40

网上的代码：

#include
#include
#include
using namespace std;
int dp[100000],sum,ans;
struct st
{
	int v;
	int m;
}data[100000];
void full(int x)
{
	for(int i=data[x].v;i<=ans;i++)
	dp[i]=max(dp[i],dp[i-data[x].v]+data[x].v);
}
void one(int x)
{
	for(int j=1;j<=data[x].m;j++)
	     for(int i=ans;i>=data[x].v;i--)
	         dp[i]=max(dp[i],dp[i-data[x].v]+data[x].v);
}
int main()
{
	int i,j,n;
	while(scanf("%d",&n)&&(n>0))
	{
		memset(dp,0,sizeof(dp));
		sum=0;
		for(i=1;i<=n;i++)
		{
			scanf("%d%d",&data[i].v,&data[i].m);
			sum+=data[i].v*data[i].m;
		}
		ans=sum/2;
		for(i=1;i<=n;i++)
		{
			if(data[i].v*data[i].m>=ans)
			full(i);
			else
			one(i);
		}
		printf("%d %d\n",sum-dp[ans],dp[ans]);
	}
	return 0;
}

//母函数方法：

/*注意将数组a,s清零，WA了好几次，测试数据都过。。无语。 
*/
#include
#include
int a[250010],s[250010];
int v[55],m[55]; 
int main()
{
	int n,i,j,k,sum,ans;
	while(scanf("%d",&n)&&n>0)
	{
		sum=0;
		memset(s,0,sizeof(s));
		memset(a,0,sizeof(a));
		for(i=1;i<=n;i++)
		{
			scanf("%d%d",&v[i],&m[i]);
			sum+=v[i]*m[i];
		}
		for(i=0;i<=v[1]*m[1];i+=v[1])//注意变化。 
		{
			s[i]=1;
		}
		for(i=2;i<=n;i++)
		{
			for(j=0;j<=sum;j++)
			{
		      	for(k=0;k+j<=sum&&k<=v[i]*m[i];k+=v[i])
				{
				   a[k+j]+=s[j];
			    }
			}
			for(k=0;k<=sum;k++)
			{
				s[k]=a[k];
				a[k]=0;
			}
		}
		for(i=sum/2;i>=0;i--)
		{
			if(s[i])
			{
			    printf("%d %d\n",sum-i,i);
			    break;
			}
		}
	}
	return 0;
}

我的代码：

#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<climits>
#include<queue>
#include<vector>
#include<map>
#include<sstream>
#include<set>
#include<stack>
using namespace std;
//typedef long long ll;
//ifstream fin("input.txt");
//ofstream fout("output.txt");
//fin.close();
//fout.close();
int val[60];
int num[60];
int n;
int dp[270000];
int main()
{
int i,j,k;
while(scanf("%d",&n)==1&&n>0)
{
int cnt=0;
for(i=1;i<=n;i++)
{
scanf("%d%d",&val[i],&num[i]);
cnt+=num[i]*val[i];
}
int sum=cnt/2;
memset(dp,0,sizeof(dp));
for(i=1;i<=n;i++)
{
for(j=1;j<=num[i];j++)
{