杭电1503（Advanced Fruits）
来源：程序员人生 发布时间：2015-05-29 07:53:03 阅读次数：2321次
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Problem Description
The company "21st Century Fruits" has specialized in creating new sorts of fruits by transferring genes from one fruit into the genome of another one. Most times this method doesn't work, but sometimes, in very rare cases, a new fruit
emerges that tastes like a mixture between both of them.
A big topic of discussion inside the company is "How should the new creations be called?" A mixture between an apple and a pear could be called an apple-pear, of course, but this doesn't sound very interesting. The boss finally decides to use the shortest string
that contains both names of the original fruits as sub-strings as the new name. For instance, "applear" contains "apple" and "pear" (APPLEar and apPlEAR), and there is no shorter string that has the same property.
A combination of a cranberry and a boysenberry would therefore be called a "boysecranberry" or a "craboysenberry", for example.
Your job is to write a program that computes such a shortest name for a combination of two given fruits. Your algorithm should be efficient, otherwise it is unlikely that it will execute in the alloted time for long fruit names.
Input
Each line of the input contains two strings that represent the names of the fruits that should be combined. All names have a maximum length of 100 and only consist of alphabetic characters.
Input is terminated by end of file.
Output
For each test case, output the shortest name of the resulting fruit on one line. If more than one shortest name is possible, any one is acceptable.
Sample Input
apple peach
ananas banana
pear peach
Sample Output
思路：先找出最大公共子序列 ，然后输出两个字符串，而最大公共子序列只输出1次。
代码实现：
import java.util.*;
class Main {
static String str1,str2;
static int[][] dp;
static int[][] mark;
public static void main(String[] args){
Scanner sc=new Scanner(System.in);
while(sc.hasNext()){
str1=sc.next();
str2=sc.next();
lcs();
printLCS(str1.length(),str2.length());
System.out.println();
}
}
public static void lcs(){
int i,j;
mark=new int[str1.length()+1][str2.length()+1];
dp=new int[str1.length()+1][str2.length()+1];
for(i=0;i<=str1.length();i++){
mark[i][0]=1;
}
for(i=0;i<=str2.length();i++){
mark[0][i]=⑴;
}
for(i=1;i<=str1.length();i++){
for(j=1;j<=str2.length();j++){
if(str1.charAt(i⑴)==str2.charAt(j⑴)){
dp[i][j]=dp[i⑴][j⑴]+1;
mark[i][j]=0;
}else if(dp[i⑴][j]>=dp[i][j⑴]){
dp[i][j]=dp[i⑴][j];
mark[i][j]=1;
}else{
dp[i][j]=dp[i][j⑴];
mark[i][j]=⑴;
}
}
}
}
public static void printLCS(int i,int j){
if(i==0&&j==0){
return ;
}
if(mark[i][j]==0){
printLCS(i⑴,j⑴);回溯输出字符
System.out.print(str1.charAt(i⑴));
}else if(mark[i][j]==1){
printLCS(i⑴,j);
System.out.print(str1.charAt(i⑴));
}else{
printLCS(i,j⑴);
System.out.print(str2.charAt(j⑴));
}
}
}
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