【搜索】HDU1016

DFS搜索，实现做预处理可以加快速度。

1.      由于0<n<20，可以先算出3~19+18的所有数是不是质数。

2.      可以事先算出每个数可以与哪些数组合，直接使用减少for循环次数

 

JAVA版当输出数据量大时不要使用System.out，使用PrintWriter。

以前用System.out一直4000MS TLE，用PrintWriter是600MS

import java.io.IOException;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.Scanner;

public class Main {
	int N;
	boolean[] appeard = new boolean[20];
	int[] path = new int[20];
	boolean[] isprime = new boolean[38];
	int[][] table = new int[20][20];
	PrintWriter out;

	public boolean isPrime(int num) {
		int sqrt = (int) Math.floor(Math.sqrt(num));
		for (int i = 2; i <= sqrt; i++) {
			if (num % i == 0) {
				return false;
			}
		}
		return true;
	}

	public static void main(String[] args) throws IOException {
		new Main().work();
	}

	public void work() throws IOException {
		Scanner input = new Scanner(System.in);
		out = new PrintWriter(new OutputStreamWriter(System.out));
		for (int i = 3; i < isprime.length; i++) {
			isprime[i] = isPrime(i);
		}
		int index;
		for (int i = 1; i < 20; i++) {
			index = 0;
			for (int j = 1; j < 20; j++) {
				if (i == j) {
					continue;
				}
				if (isprime[i + j]) {
					table[i][index++] = j;
				}
			}
		}
		appeard[1] = true;
		path[1] = 1;
		int counter = 1;
		while (input.hasNext()) {
			N = input.nextInt();
			input.nextLine();
			out.println("Case " + counter++ + ":");
			dfs1(1);
			out.println();
			out.flush();
		}
		input.close();
	}

	public void dfs1(int alreadyAppeardCount) throws IOException {
		int previous = path[alreadyAppeardCount];
		int next;
		if (alreadyAppeardCount == N) {
			if (isprime[previous + 1]) {//*//
				for (int i = 1; i <= N; i++) {
					out.print(path[i]);
					if (i != N) {
						out.print(" ");
					} else {
						out.println();
					}
				}
			}
		}
		for (int i = 0; i <= N; i++) {
			next = table[previous][i];
			if (next == 0) {
				return;
			}
			if (next > N) {
				continue;
			}
			if (appeard[next]) {
				continue;
			}
			appeard[next] = true;//*//
			path[alreadyAppeardCount + 1] = next;
			dfs1(alreadyAppeardCount + 1);//*//
			appeard[next] = false;//*//
		}
	}
}