hdu1171 Big Event in HDU --01背包

最新推荐文章于 2020-04-11 15:07:56 发布

LaoJiu_

最新推荐文章于 2020-04-11 15:07:56 发布

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分类专栏：【动态规划】--背包问题

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本文针对一个具体的设施分配问题，提供了一种通过动态规划求解最优解的方法。问题要求将不同种类和数量的设施等价地分配给两个学院，并确保两者的总价值尽可能接近且前者不低于后者。

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原题链接： http://acm.hdu.edu.cn/showproblem.php?pid=1171

一：原题内容

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

Sample Output

二：分析理解

直接看代码就懂了。

三：AC代码

#include<iostream>  
#include<algorithm>  

using namespace std;

int va[5005];
int dp[250005];
int a, b;
int n;

int main()
{
	while (~scanf("%d", &n) && n > 0)
	{
		memset(dp, 0, sizeof(dp));

		int sum = 0;
		int num = 1;
		for (int i = 1; i <= n; i++)
		{
			scanf("%d%d", &a, &b);
			while (b--)
			{
				va[num++] = a;
				sum += a;
			}
		}

		for (int i = 1; i < num; i++)
			for (int j = sum / 2; j >= va[i]; j--)
				dp[j] = max(dp[j - va[i]] + va[i], dp[j]);

		printf("%d %d\n", sum - dp[sum / 2], dp[sum / 2]);
	}

	return 0;
}