HDU1171 - Big Event in HDU 动态规划之多重背包

1.题目叙述：

Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 38791    Accepted Submission(s): 13417

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

 

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 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

 

Author

lcy

 

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2.题目描述：

给出每个物体的价值和物体的数量，如何分使得A,B所得价值最接近并且A的价值不能小于B。

3.解题思路：

要使得价值最相近相当于有两个背包重量分别是sum / 2分别存sum - dp[sum / 2] 和dp[sum / 2]件物品的价值嘛，多重背包问题的话转换成二进制解决。

ps：然后这题还有一个解法是母函数，因为集训时候没学，以后学了补上orz

4.AC代码：

#include 
#include 
#include 
#define maxn 202000
#define N 55
using namespace std;
int v[N], w[N], dp[maxn];

int main()
{
	int n;
	while (scanf("%d", &n) != EOF && n >= 0)
	{
		memset(dp, 0, sizeof(dp));
		int sum = 0;
		for (int i = 0; i < n; i++)
		{
			scanf("%d%d", &v[i], &w[i]);
			sum += v[i] * w[i];
		}
		for (int i = 0; i < n; i++)
			for (int j = 0; j < w[i]; j++)
				for (int k = sum / 2; k >= v[i]; k--)
					dp[k] = max(dp[k], dp[k - v[i]] + v[i]);
		printf("%d %d\n", sum - dp[sum / 2], dp[sum / 2]);
	}
    return 0;
}