HDU 1171 Big Event in HDU （多重背包）

Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19108    Accepted Submission(s): 6707

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

2 10 1 20 1 3 10 1 20 2 30 1 -1

 

 

Sample Output

20 10 40 40

 

 

题意：学校分为计算机学院，和软件学院，已知有N种价值的设备，每种价值为v，有m件，现将设备平分到两个学院当中，使每部分价值尽量相等（如果不等则使第一部分大于第二部分）。

 

多重背包：

思路：背包体积为总价值的一半，然后用多重背包求最大值。

01背包，完全背包，多重背包 ，模板代码：http://blog.csdn.net/deng_hui_long/article/details/10603015

import java.io.*;
import java.util.*;
public class Main {
	int n;
	int dp[]=new int[100000];
	int sum,vsum;
	public static void main(String[] args) {
		new Main().work();
	}
	void work(){
		Scanner sc=new Scanner(new BufferedInputStream(System.in));
		while(sc.hasNext()){
			n=sc.nextInt();
			if(n<0) break;
			Node node[]=new Node[n];
			Arrays.fill(dp, 0);
			vsum=sum=0;
			for(int i=0;i<n;i++){
				int v=sc.nextInt();
				int m=sc.nextInt();
				sum+=v*m;
				node[i]=new Node(v,m);
			}
			vsum=sum>>1;// 设备要尽量评分，所以要除以2：注：右一位代表除以2
			for(int i=0;i<n;i++){
				multiplyPack(node[i].v,node[i].v,node[i].m);
			}
			
			System.out.println((sum-dp[vsum])+" "+dp[vsum]);
		}
	}
	void multiplyPack(int cost,int weight,int amount){
		if(cost*amount>=vsum){//如果价值大于总共价值的一半，假设设备是无限的,按照完全背包来处理
			completePack(cost,weight);
		}
		else{//如果小于总共价值的一半，即按照01背包来处理
			int k=1;
			while(k<amount){
				zeroOnePack(k*cost,k*weight);
				amount-=k;
				k<<=1;//左一位代表乘以2
			}
			zeroOnePack(amount*cost,amount*weight);
		}
	}
	//01背包
	void zeroOnePack(int cost,int weight){
		for(int i=vsum;i>=cost;i--){
			dp[i]=Math.max(dp[i],dp[i-cost]+weight);
		}
	}
	//完全背包
	void completePack(int cost,int weight){
		for(int i=cost;i<=vsum;i++){
			dp[i]=Math.max(dp[i],dp[i-cost]+weight);
		}
	}
	class Node{
		int v;
		int m;
		Node(int v,int m){
			this.v=v;
			this.m=m;
		}
	}
}