ural 1005. Stone Pile dp

本文介绍了一个经典的计算机科学问题:如何将已知重量的石头分成两堆,使两堆之间的重量差最小。通过动态规划算法求解,并提供了一个Java实现的例子。

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1005. Stone Pile

Time limit: 1.0 second
Memory limit: 64 MB
You have a number of stones with known weights w1, …, wn. Write a program that will rearrange the stones into two piles such that weight difference between the piles is minimal.

Input

Input contains the number of stones n (1 ≤ n ≤ 20) and weights of the stones w1, …, wn (integers, 1 ≤ wi ≤ 100000) delimited by white spaces.

Output

Your program should output a number representing the minimal possible weight difference between stone piles.

Sample

input output
5
5 8 13 27 14
3
Problem Source: USU Championship 1997
import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Main {

	public static void main(String[] args) {
		new Task().solve();
	}
}

class Task {
	InputReader in = new InputReader(System.in);
	PrintWriter out = new PrintWriter(System.out);

	void solve() {
		while(in.hasNext()){
			int n = in.nextInt() ;
			int[] nums = in.nextInts(n) ;
		    int sum = 0 ;
		    for(int i = 0 ; i < n ; i++){
		    	sum += nums[i] ;
		    }
		    boolean[] dp = new boolean[sum+1] ;
		    Arrays.fill(dp , false) ;
		    dp[0] = true ;
		    for(int i = 0 ; i < n ; i++){
		    	for(int j = sum ; j >= nums[i] ; j--){
		    		if(dp[j - nums[i]]){
		    			dp[j] = true ;
		    		}
		    	}
		    }
		    int result = Integer.MAX_VALUE ; 
		    for(int i = 0  ; i <= sum ; i++){
		    	if(dp[i]){
		    		result = Math.min(result , Math.abs(i - (sum-i))) ;
		    	}
		    }
		    out.println(result) ;
		}
		out.flush();
	}
	
}

class InputReader {
	public BufferedReader reader;
	public StringTokenizer tokenizer;

	public InputReader(InputStream stream) {
 		reader = new BufferedReader(new InputStreamReader(stream), 32768);
		tokenizer = new StringTokenizer("");
	}

	private void eat(String s) {
		tokenizer = new StringTokenizer(s);
	}

	public String nextLine() {
		try {
			return reader.readLine();
		} catch (Exception e) {
			return null;
		}
	}

	public boolean hasNext() {
		while (!tokenizer.hasMoreTokens()) {
			String s = nextLine();
			if (s == null)
				return false;
			eat(s);
		}
		return true;
	}

	public String next() {
		hasNext();
		return tokenizer.nextToken();
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}
	
	public int[] nextInts(int n){
		int[] nums = new int[n] ;
		for(int i = 0 ; i < n ; i++){
			nums[i] = nextInt() ;
		}
		return nums ;
	}

	public long nextLong() {
		return Long.parseLong(next());
	}

	public double nextDouble() {
		return Double.parseDouble(next());
	}

	public BigInteger nextBigInteger() {
		return new BigInteger(next());
	}

}



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