hdu1171 Big Event in HDU

最新推荐文章于 2020-04-11 15:07:56 发布

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本文介绍了一种资源分配算法，用于将不同价值的设施等分为两个学院，确保两者的总价值尽可能相等且A组价值不低于B组。通过使用母函数求解所有可能组合，并从总价值一半开始查找符合条件的第一个解决方案。

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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

20 10
40 40
题意：给出n个资源，每个资源包含价值及其个数，将其分为两组，使得每组的价值尽量一样，同时保证a组>=b组
思路：用母函数先求出所有可能的组合，然后从总价值的一半开始搜索，第一个满足的就是。
代码如下：
import java.io.FileInputStream;
import java.io.OutputStreamWriter;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Scanner;
import java.util.Arrays;

public class Main implements Runnable
{  
    private static final boolean DEBUG = false;
    private static final int N = 51;
    private static final int MAX = 250001;
    private int n;
    private int[] v = new int[N];
    private int[] m = new int[N];
    private Scanner cin;
    private PrintWriter cout;
    private int total;
    private boolean[] ans = new boolean[MAX];
    
    private void init()
    {
        try {
            if (DEBUG) {
                cin = new Scanner(new InputStreamReader(new FileInputStream("f:\\OJ\\uva_in.txt")));
            } else {
                cin = new Scanner(new InputStreamReader(System.in));
            }
            
            cout = new PrintWriter(new OutputStreamWriter(System.out));
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
    
    private boolean input()
    {
        n = cin.nextInt();
        if (n < 0) return false;
        
        total = 0;
        for (int i = 0; i < n; i++) {
            v[i] = cin.nextInt();
            m[i] = cin.nextInt();
            total += v[i] * m[i];
        }
        return true;
    }
    
    private void work(int limit)
    {
        Arrays.fill(ans, false);
        for (int i = 0; i <= m[0]; i++) {
            ans[i * v[0]] = true;
        }
        
        for (int i = 1; i < n; i++) {
            for (int j = 0; j <= limit; j++) {
                for (int k = 0; k <= m[i] && j + k * v[i] <= limit; k++) {
                    if (ans[j]) {
                        ans[j + k * v[i]] = true;
                    }
                }
            }
        }
    }
    
    private void solve()
    {
        work((total + 1) / 2);
        
        int a = 0, b = 0;
        
        for (int i = (total + 1) / 2; i >= 0; i--) {
            if (ans[i]) {
                a = i;
                break;
            }
        }
        
        b = total - a;
        a = Math.max(a, b);
        b = total - a;
        
        cout.println(a + " " + b);
        cout.flush();
    }
    
    @Override
    public void run()
    {
        init();
        while (input()) 
        {
            solve();
        }
    }
    
    public static void main(String[] args) 
    {
        // TODO code application logic here
       new Thread(new Main()).start();
    }
}