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UVA 10069 Distinct Subsequences(DP)
考虑两个字符串,我们用dp[i][j]表示字串第到i个和字符串到第j个的总数,因为字串必须连续
因此dp[i][j]可以有dp[i][j-1]和dp[i-1][j-1]递推而来,而不能由dp[i-1][j]递推而来。而后者的条件
是字串的第i个和字符串相等。
A subsequence of a given sequence is just the given sequence with some elements (possibly none) left out. Formally, given a sequence X =x1x2…xm, another sequence Z = z1z2…zk is a subsequence of X if there exists a strictly increasing sequence <i1, i2, …, ik> of indices of Xsuch that for all j = 1, 2, …, k, we have xij = zj. For example, Z = bcdb is a subsequence of X = abcbdab with corresponding index sequence< 2, 3, 5, 7 >.
In this problem your job is to write a program that counts the number of occurrences of Z in X as a subsequence such that each has a distinct index sequence.
Input
The first line of the input contains an integer N indicating the number of test cases to follow.
The first line of each test case contains a string X, composed entirely of lowercase alphabetic characters and having length no greater than 10,000. The second line contains another string Z having length no greater than 100 and also composed of only lowercase alphabetic characters. Be assured that neither Z nor any prefix or suffix of Z will have more than 10100 distinct occurrences in X as a subsequence.
Output
For each test case in the input output the number of distinct occurrences of Z in X as a subsequence. Output for each input set must be on a separate line.
Sample Input
2babgbag
bag
rabbbit
rabbit
Sample Output
53
import java.io.*; import java.math.*; import java.util.*; public class Main{ public static void main(String []args){ Scanner cin=new Scanner(System.in); int t=cin.nextInt(); while(t--!=0){ char a[]=cin.next().toCharArray(); char b[]=cin.next().toCharArray(); // System.out.println("2333 "); BigInteger [][] dp=new BigInteger[110][10100]; for(int i=0;i<dp.length;i++){ for(int j=0;j<dp[i].length;j++) dp[i][j]=BigInteger.ZERO; } // System.out.println("2333 "); for(int j=0;j<a.length;j++){ if(j>0) dp[0][j]=dp[0][j-1]; if(b[0]==a[j]) dp[0][j]=dp[0][j].add(BigInteger.ONE); } // System.out.println("2333 "); for(int i=1;i<b.length;i++){ for(int j=i;j<a.length;j++){ dp[i][j]=dp[i][j-1]; if(b[i]==a[j]) dp[i][j]=dp[i][j].add(dp[i-1][j-1]); } } System.out.println(dp[b.length-1][a.length-1]); } } }
UVA 10069 Distinct Subsequences(DP)