hdoj 1503 Advanced Fruits (LCS 变形合并)

Advanced Fruits

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2995    Accepted Submission(s): 1531
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

 

Source

University of Ulm Local Contest 1999

 

题目大意：两个字符串输出  其公共部分合并只输出一次

思路：把动态规划状态转移方程遍历到的状态进行标记，从大到小遍历dp[ i ][ j ]路径，从小到大回溯输出

代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;char a[110],b[110];//输入字符串 int mark[110][110],dp[110][110];//标记状态  //记录路径 int len1,len2;void LCS(){int i,j;memset(dp,0,sizeof(dp));for(i=0;i<=len1;i++)mark[i][0]=1;for(j=0;j<=len2;j++)mark[0][j]=-1;for(i=1;i<=len1;i++){for(j=1;j<=len2;j++){if(a[i-1]==b[j-1]){dp[i][j]=dp[i-1][j-1]+1;mark[i][j]=0; }  else if(dp[i-1][j]>=dp[i][j-1]) { dp[i][j]=dp[i-1][j]; mark[i][j]=1; } else { dp[i][j]=dp[i][j-1]; mark[i][j]=-1; }}} }   void out(int i,int j)//遍历路径  回溯输出  { if(!i&&!j) return ; if(mark[i][j]==0) { out(i-1,j-1); printf("%c",a[i-1]); } else if(mark[i][j]==1) { out(i-1,j); printf("%c",a[i-1]); } else { out(i,j-1); printf("%c",b[j-1]); } }int main(){while(~scanf("%s%s",&a,&b)){len1=strlen(a);len2=strlen(b);LCS();out(len1,len2);printf("\n");}return 0; } 

做了这道题，对LCS的状态转移方程又有了更清晰的认识，求LCS时他的路径也是唯一的