F - 6 HDU - 1171 Big Event in HDU

探讨了杭州电子科技大学计算机学院在2002年拆分为计算机学院和软件学院时,如何公平分配设施的问题。通过算法求解最优分配方案,确保两学院获得的设施价值尽可能相等。

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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

问题链接:https://vjudge.net/contest/279611#problem/F

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn = 1000010;
int a[maxn], b[maxn];
int dp[maxn];
bool cmp(int a, int b)
{
	return a>b;
}
int main()
{
	int n;
	while (cin >> n && n>0)
	{
		int sum = 0;
		memset(dp, 0, sizeof(dp));
		memset(a, 0, sizeof(a));
		int cnt = 0;
		while (n--)
		{
			int x, y;
			cin >> x >> y;
			sum += x * y;
			while (y--)
			{
				a[cnt++] = x;
			}
		}
		sort(a, a + cnt, cmp);
		for (int i = 0; i<cnt; i++)
		{
			for (int j = sum / 2; j >= a[i]; j--)
			{
				dp[j] = max(dp[j], dp[j - a[i]] + a[i]);
			}
		}
		cout << sum - dp[sum / 2] << " " << dp[sum / 2] << endl;
	}
	return 0;
}
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