Poj 1936,3302 Subsequence(LCS)

一、Description（3302）

Given a string s of length n, a subsequence of it, is defined as another string
s' = s_u1s_u2...s_um where 1 ≤
u₁ < u₂ < ... < u_m ≤
n and s_i is the ith character of s. Your task is to write a program that, given two strings
s1 and s2, checks whether either s2 or its reverse is a subsequence of
s1 or not.

Input

The first line of input contains an integer T, which is the number of test cases. Each of the next
T lines contains two non-empty strings s1 and s2 (with length at most 100) consisted of only alpha-numeric characters and separated from each other by a single space.

Output

For each test case, your program must output "YES", in a single line, if either
s2 or its reverse is a subsequence of s1. Otherwise your program should write "NO".

一、Description（1936）

You have devised a new encryption technique which encodes a message by inserting between its characters randomly generated strings in a clever way. Because of pending patent issues we will not discuss in detail how the strings
are generated and inserted into the original message. To validate your method, however, it is necessary to write a program that checks if the message is really encoded in the final string.




Given two strings s and t, you have to decide whether s is a subsequence of t, i.e. if you can remove characters from t such that the concatenation of the remaining characters is s.

Input

The input contains several testcases. Each is specified by two strings s, t of alphanumeric ASCII characters separated by whitespace.The length of s and t will no more than 100000.

Output

For each test case output "Yes", if s is a subsequence of t,otherwise output "No".

二、题解

        这两道题目都是求LCS问题，只是求解的形式有所不同。求的不是LCS而是判断是否为子序列，其中3302还求逆序列是否为子序列。但不管怎么变，只要记住DP求LCS算法，题目就很容易求解。

三、java代码

    import java.util.Scanner;     

    public class Main {
        public static int LCS(String x,String y){
            int  [][] z=new int [x.length()+1][y.length()+1];
            int i,j;
            for( i=0;i<=x.length();i++)
                z[i][0]=0;
            for( j=0;j<=y.length();j++)
                z[0][j]=0;    

            for(i=1;i<=x.length();i++){
                for( j=1;j<=y.length();j++){
                    if(x.charAt(i-1)==y.charAt(j-1)){
                        z[i][j]= z[i-1][j-1]+1;
                    }
                    else
                        z[i][j]=z[i-1][j] > z[i][j-1] ?z[i-1][j]:z[i][j-1];
                }
            }
            return z[x.length()][y.length()];
        }
        public static void main(String[] args) {
           Scanner cin = new Scanner(System.in);
           String s,s1;
           int n,i;
           n=cin.nextInt();
           for(i=0;i<n;i++){
        	   s=cin.next();
        	   s1=cin.next();
	          if( LCS(s,s1)==s1.length() || LCS(s,new StringBuffer(s1).reverse().toString())==s1.length()){
	        	  System.out.println("YES");
	          }else{
	        	  System.out.println("NO");
	          }
           }
     }
 }     

版权声明：本文为博主原创文章，未经博主允许不得转载。