Big Event in HDU HDU 1171 多重背包

最新推荐文章于 2020-04-11 15:07:56 发布

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本文介绍了一个关于设施在计算机学院和软件学院之间的分配问题。通过一个具体的案例，文章详细阐述了如何利用动态规划的方法来解决这一问题，确保两个学院获得的设施价值尽可能相等且计算机学院的设施价值不低于软件学院。

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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

Sample Output

20 1040 40

代码：

#include <iostream>
#include<stdio.h>
#include<string.h>

using namespace std;

#define INF 0x3f3f3f3f
int v[100005];
int dp[100005];
int w[100];
int main()
{
	int n,sum,num;
	while(~scanf("%d",&n))
	{
		sum=0;
		if(n<=0) break;
		memset(dp,0,sizeof(dp));
		for(int i=0;i<n;i++)
		{
			scanf("%d%d",&v[i],&w[i]);
			sum+=(v[i]*w[i]);
		}
		num=sum/2;
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<w[i];j++)
			{
				for(int k=num;k>=v[i];k--)
				{
					 dp[k]=max(dp[k],dp[k-v[i]]+v[i]);
				}
		              
			}
		       
		}
		printf("%d %d\n",sum-dp[num],dp[num]);
		   
	}
	
}