day3 HDU - 1171

最新推荐文章于 2020-07-24 21:13:21 发布

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最新推荐文章于 2020-07-24 21:13:21 发布

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本文介绍了一个关于设施价值在两个学院间进行尽可能平均分配的问题。通过动态规划算法，确保了计算机学院和软件学院获得的设施价值尽可能相等，同时保证前者不低于后者。详细解析了输入输出格式，以及算法实现。

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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
2
10 1
20 1
3
10 1
20 2
30 1
-1
Sample Output
20 10
40 40
题目大意就是尽可能的平均分，并且前一个数要大于后一个数；
dp[i]表示最接近i这个数的最大数；

#include <iostream>
using namespace std;
#include<cmath>
int dp[125060];
int a[120000],m;
#include<bits/stdc++.h>
int main()
{

	int n;
	while(cin>>n&&n>0)
	{
		memset(dp,0,sizeof(dp));
		int t=0;
		int sum=0,aa;
		while(n--)
		{
			cin>>aa>>m;
			sum+=aa*m;
			for(int i=1;i<=m;i++)
			{
				a[t++]=aa;
			}
		}
		
		int k=sum;
		k/=2;
		for(int i=0;i<t;i++)
		{
			for(int j=k;j>=a[i];j--)
			{
				dp[j]=max(dp[j],dp[j-a[i]]+a[i]);
			}
		}
		cout<<sum-dp[k]<<" "<<dp[k]<<endl;
	}
}