HDU1171 Big Event in HDU（多重背包）

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本文介绍了一种解决特定多重背包问题的方法，通过二进制优化技术来提高计算效率。该问题要求将一系列设施（每种有固定价值和数量）平均分配给两个学院，使两者的总价值尽可能接近。

题目描述

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 10
40 40

分析

根据分析，本题是n件物品，对应价值v和个数m，求一半总价值的最大值，是多重背包问题。
可以使用二进制优化多重背包加快运算速度。

C++ 代码

#include<bits/stdc++.h>
using namespace std;
const int N = 1e4+10;
int f[N*10],w[N],n,a,c;
int main()
{
	ios::sync_with_stdio(false);
	while(cin>>n)
	{
		if(n<=0) break;
		int sum=0,co=0;
		memset(f,0,sizeof f);
		for(int i=0;i<n;i++)
		{
			cin>>a>>c;
			sum+=(c*a);
			for(int k=1;k<=c;k*=2)    //每个物品划分为二进制形式  
			{
				w[co++]=a*k;
				c-=k;
			}
			if(c) w[co++]=a*c;
		}
		int v=sum/2;                //要求一半的体积下的最大值
		for(int i=0;i<co;i++)       //二进制优化后就变成了0-1背包问题
		{
			for(int j=v;j>=w[i];j--)
				f[j]=max(f[j],f[j-w[i]]+w[i]);
		}
		cout<<max(sum-f[v],f[v])<<" "<<min(sum-f[v],f[v])<<endl;    //输出此时的最大值和最小值
	}
	return 0;
}