HDU 1297 Children’s Queue

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1297

题面：

Children’s Queue Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13363    Accepted Submission(s): 4379


Problem Description There are many students in PHT School. One day, the headmaster whose name is PigHeader wanted all students stand in a line. He prescribed that girl can not be in single. In other words, either no girl in the queue or more than one girl stands side by side. The case n=4 (n is the number of children) is like
FFFF, FFFM, MFFF, FFMM, MFFM, MMFF, MMMM
Here F stands for a girl and M stands for a boy. The total number of queue satisfied the headmaster’s needs is 7. Can you make a program to find the total number of queue with n children?
 
Input There are multiple cases in this problem and ended by the EOF. In each case, there is only one integer n means the number of children (1<=n<=1000)  
Output For each test case, there is only one integer means the number of queue satisfied the headmaster’s needs.  
Sample Input

1 2 3  
Sample Output

1 2 4  
Author SmallBeer (CML)  
题目大意：     给定队伍长度，求不出现单独一个女生的方案数。
解题：     可以用dp三维来解决这个问题。dp[i][j][k],i表示是第i位，j为1表示男生，j为0表示女生，k为0表示0个女生，k为1表示1个女生，k为2表示多个女生，即合法状态。     递推关系如下：      

dp[i+1][0][2]=dp[i+1][0][2].add(dp[i][0][1]);
dp[i+1][0][2]=dp[i+1][0][2].add(dp[i][0][2]);
dp[i+1][1][0]=dp[i+1][1][0].add(dp[i][0][2]);
dp[i+1][1][0]=dp[i+1][1][0].add(dp[i][1][0]);
dp[i+1][0][1]=dp[i+1][0][1].add(dp[i][1][0]);

部分状态虽然不符合最后的要求，即女生不可落单，但计算过程中需要用到相应值，最后的答案为，dp[n][1][0]+dp[n][0][2]

代码：

import java.io.*;
import java.util.*;
import java.math.*;
public class Main{
	public static void main(String args[])
	{
		BigInteger dp[][][]=new BigInteger [1005][2][3];
		Scanner sc =new Scanner(new BufferedInputStream(System.in));
		PrintWriter cout=new PrintWriter(System.out);
		for(int i=0;i<=1004;i++)
			for(int j=0;j<=1;j++)
				for(int k=0;k<=2;k++)
					dp[i][j][k]=BigInteger.ZERO;
		dp[1][0][1]=BigInteger.valueOf(1);
		dp[1][1][0]=BigInteger.valueOf(1);
		for(int i=1;i<=1000;i++)
		{
			dp[i+1][0][2]=dp[i+1][0][2].add(dp[i][0][1]);
			dp[i+1][0][2]=dp[i+1][0][2].add(dp[i][0][2]);
			dp[i+1][1][0]=dp[i+1][1][0].add(dp[i][0][2]);
			dp[i+1][1][0]=dp[i+1][1][0].add(dp[i][1][0]);
			dp[i+1][0][1]=dp[i+1][0][1].add(dp[i][1][0]);
		}
		int t;
		while(sc.hasNext())
		{
           t=sc.nextInt();
		   cout.println(dp[t][0][2].add(dp[t][1][0]));
		}
		cout.flush();
	}
}