HDOJ 1503 Advanced Fruits(LCS+记录路径)

Advanced Fruits

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2136    Accepted Submission(s): 1089
Special Judge

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 peachananas bananapear peach

 

Sample Output

appleachbananaspearch题目大意：两个字符串，求出将最长公共子序列去下一个之后组成的那个串思路：两个串求出最长公共子序列之后标记一下，然后存入数组（存一次），然后其它字符都要存入数组最后倒序输出ac代码：
#include<stdio.h>#include<string.h>#include<math.h>#include<iostream>#include<algorithm>#define MAXN 110#define MAX(a,b) a>b?a:busing namespace std;char s1[MAXN];char s2[MAXN];int v[MAXN];char ans[MAXN];int dp[110][110];int main(){int i,j;while(scanf("%s%s",s1+1,s2+1)!=EOF)//刚开始从0开始写，但是写着写着发现乱了，然后又滚回来从1写{int len1=strlen(s1+1);int len2=strlen(s2+1);memset(dp,0,sizeof(dp));for(i=1;i<=len1;i++){for(j=1;j<=len2;j++){dp[i][j]=s1[i]==s2[j]?(dp[i-1][j-1]+1):max(dp[i-1][j],dp[i][j-1]);}}int cnt=0;while(dp[len1][len2]){if(s1[len1]==s2[len2]){ans[cnt++]=s1[len1];len1--;len2--;}elsedp[len1-1][len2]>dp[len1][len2-1]?ans[cnt++]=s1[len1--]:ans[cnt++]=s2[len2--];}while(len1>0)ans[cnt++]=s1[len1--];while(len2>0)ans[cnt++]=s2[len2--];for(i=cnt-1;i>=0;i--)printf("%c",ans[i]);printf("\n");}return 0;}