LightOJ-1013 LCS+递推

1013 - Love Calculator

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Time Limit: 2 second(s)

Memory Limit: 32 MB

Yes, you are developing a 'Love calculator'. The software would be quite complex such that nobody could crack the exact behavior of the software.

So, given two names your software will generate the percentage of their 'love' according to their names. The software requires the following things:

1.                  The length of the shortest string that contains the names as subsequence.

2.                   Total number of unique shortest strings which contain the names as subsequence.

Now your task is to find these parts.

Input

Input starts with an integer T (≤ 125), denoting the number of test cases.

Each of the test cases consists of two lines each containing a name. The names will contain no more than 30 capital letters.

Output

For each of the test cases, you need to print one line of output. The output for each test case starts with the test case number, followed by the shortest length of the string and the number of unique strings that satisfies the given conditions.

You can assume that the number of unique strings will always be less than 2⁶³. Look at the sample output for the exact format.

Sample Input	Output for Sample Input
3 USA USSR LAILI MAJNU SHAHJAHAN MOMTAJ	Case 1: 5 3 Case 2: 9 40 Case 3: 13 15

 

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PROBLEM SETTER: JANE ALAM JAN

题目链接：

http://lightoj.com/volume_showproblem.php?problem=1013

题目大意：

  给出两个字符串A、B，求一个包含这两个字符串为子序列的字符串的最小长度，并求出方案数。

解题思路：

  最小长度可以通过求最长公共子序列得到，为len(A)+len(B)-LCS(A,B)。

  难点在于得到方案数，令f[i][j]为取A前i个字符，B前j个字符得到最短字符串的方案数，

则有 dp[i][j-1]>dp[i-1][j]时，f[i][j]=f[i][j-1];

   dp[i][j-1]<dp[i-1][j]时，f[i][j]=f[i-1][j];

   dp[i][j-1]=dp[i-1][j]时，f[i][j]=f[i][j-1]+f[i-1][j]。

  网上还有另一种递推方式，是用f[k][i][j]表示取A前i个字符，B前j个字符构造长度为k的字符串的方案数。 但递推式为

  A[i]==B[j]时，f[k][i][j]=f[k-1][i][j];

  A[i]!=B[j]时，f[k][i][j]=f[k-1][i][j-1]+f[k-1][i-1][j]。

  这样递推对于构造最短串来说方案数是没有问题的，但是非最短串方案数少。

  举例来说，两个字符串AA、DA，f[4][2][2]=0就不正确，但f[3][2][2]=2却是正确的，这是因为f[4][2][2]的上一个状态

f[3][1][1]=0，所以这种递推的定义并不明确。

AC代码：

import java.util.*;

public class Main {
	static int T,n,m,ans;
	static char[] aa,bb;
	static int[][] dp=new int[35][35];
	static long[][] ff=new long[35][35];

	public static void main(String[] args) {
		Scanner in=new Scanner(System.in);
		T=in.nextInt();
		for(int t=1;t<=T;t++)
		{
			aa=(" "+in.next()).toCharArray();
			bb=(" "+in.next()).toCharArray();
			n=aa.length-1;m=bb.length-1;
			for(int i=0;i<=n;i++) ff[i][0]=1;
			for(int i=0;i<=m;i++) ff[0][i]=1;
			for(int i=1;i<=n;i++)
			for(int j=1;j<=m;j++)
			{
				if(aa[i]==bb[j])
				{
					dp[i][j]=dp[i-1][j-1]+1;
					ff[i][j]=ff[i-1][j-1];
				}
				else
				{
					dp[i][j]=Math.max(dp[i-1][j],dp[i][j-1]);
					ff[i][j]=0;
					if(dp[i][j-1]>=dp[i-1][j])
						ff[i][j]+=ff[i][j-1];
					if(dp[i][j-1]<=dp[i-1][j])
						ff[i][j]+=ff[i-1][j];
				}
			}
			ans=n+m-dp[n][m];
			System.out.println("Case "+t+": "+ans+" "+ff[n][m]);
		}
	}
}