ACM超时，help

[qq_43164599]

Sample Input

3 3 //两数组的长度
1 2 1 //第一个数组
2 1 1 //第二个数组
1 2 1 1 2 1 //合数组
3 3
1 2 1
2 1 2
1 1 1 2 2 2
0 0

Sample Output

possible
not possible

题目的意思： 给你两个数组a,b 让a,b两个数组按其原序（单个数组内的元素前后顺序不变，第二个数组的元素可以插入到第一个里面）进行组合，问能否组合成为c数组？

//////////代码超时了，dfs究竟有什么问题呢？

package dynamicProgramming;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class LISTest {

public static int N1, N2;
public static int[] train1 = new int[1001];
public static int[] train2 = new int[1001];
public static int[] combain = new int[2002];
public static boolean flag;
public static void main(String[] args) throws IOException {
BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer sTokenizer = null;
sTokenizer = new StringTokenizer(input.readLine());
//FileInputStream fileInputStream= new FileInputStream("C:\\Users\\LAC\\Desktop\\新建文件夹 (2)\\input.txt");
N1 = Integer.parseInt(sTokenizer.nextToken());
N2 = Integer.parseInt(sTokenizer.nextToken());
while(N1 != 0 && N2 != 0) {
sTokenizer = new StringTokenizer(input.readLine());
for(int i = 1; i <= N1; i++) train1[i] = Integer.parseInt(sTokenizer.nextToken());
sTokenizer = new StringTokenizer(input.readLine());
for(int j = 1; j <= N2; j++) train2[j] = Integer.parseInt(sTokenizer.nextToken());
sTokenizer = new StringTokenizer(input.readLine());
for(int k = 1; k <= N1+N2; k++) combain[k] = Integer.parseInt(sTokenizer.nextToken());
flag = false;
dfs(1, 1, 1);
if(flag) System.out.print("possible");
else System.out.print("not possible");
sTokenizer = new StringTokenizer(input.readLine());
N1 = Integer.parseInt(sTokenizer.nextToken());
N2 = Integer.parseInt(sTokenizer.nextToken());
if(N1 != 0 && N2 != 0) System.out.println();
}
}
public static void dfs(int i, int j, int n) { //i，j表示当前结点的左右子结点在train1和train2中的下标
if(n == N1 + N2+1) {
flag = true;
return;
}
if(flag == false && i <= N1 && train1[i] == combain[n]) dfs(i+1, j, n+1);
if(flag == false && j <= N2 && train2[j] == combain[n]) dfs(i, j+1, n+1);
}

}

[qq_43164599]

[qq_43164599]

hdu3779

[qq_39936465]

StringTokenizer现在应该不用了把，已被string.split 替代了。

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class test12 {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		String s1[], s2[], s3[], length[];
		List<Array> list = new ArrayList<Array>();
		Scanner sc = new Scanner(System.in);
		do {
			System.out.println("请输入2个数组的长度用空格分开：");
			length = sc.nextLine().split(" ");
//			 数组长度为0时 结束输入  
			if (Integer.parseInt(length[0]) == 0 || Integer.parseInt(length[1]) == 0) {
				break;
			}
			do {
				System.out.println("请输入第一个数组数字用空格分开：");
				s1 = sc.nextLine().split(" ");
			} while (s1.length != Integer.parseInt(length[0]));
			do {
				System.out.println("请输入第二个数组数字用空格分开：");
				s2 = sc.nextLine().split(" ");
			} while (s2.length != Integer.parseInt(length[1]));
			System.out.println("请输入组合数组数字用空格分开：");
			s3 = sc.nextLine().split(" ");
			Array arr = new Array(s1, s2, s3);
			list.add(arr);
		} while (true);
		for (Array i : list) {
			if (i.s3.length != (i.s1.length + i.s2.length)) {
				break;
			} else {
				i.combain(0, 0, 0);
			}
			if (i.isFlag()) {
				System.out.println("Possible");
			} else {
				System.out.println("Not possible");
			}
		}

	}

}

// 建立一个3个字符串数组加一个布尔判断符的类 
class Array {
	String[] s1, s2, s3;
	boolean flag = false;

	public boolean isFlag() {
		return flag;
	}

	public void setFlag(boolean flag) {
		this.flag = flag;
	}

	public String[] getS1() {
		return s1;
	}

	public void setS1(String[] s1) {
		this.s1 = s1;
	}

	public String[] getS2() {
		return s2;
	}

	public void setS2(String[] s2) {
		this.s2 = s2;
	}

	public String[] getS3() {
		return s3;
	}

	public void setS3(String[] s3) {
		this.s3 = s3;
	}

	public Array(String[] s1, String[] s2, String[] s3) {
		super();
		this.s1 = s1;
		this.s2 = s2;
		this.s3 = s3;
	}

//	s1和s2 能否合并为s3 的判断方法 
	public void combain(int a, int b, int c) {
		if (a < s1.length) {
			if (s3[c].equals(s1[a]) && c <= s3.length) {
				if (s1.length == a + 1 && s2.length == b && s3.length == c + 1) {
					setFlag(true);
				} else {
					combain(a + 1, b, c + 1);
				}
			}
		}
		if (b < s2.length) {
			if (s3[c].equals(s1[b]) && c < s3.length) {
				if (s1.length == a && s2.length == b + 1 && s3.length == c + 1) {
					setFlag(true);
				} else {
					combain(a, b + 1, c + 1);
				}
			}
		}
	}
}