hdu1503:advanced Fruits "LCS"

Advanced Fruits Time Limit:2000/1000ms (java/other) Memory limit:65536/32768k (Java/other) total submission (s): 5 Accepted Submission (s): 2Special Judge problem DescriptionThe Company "21st Century Fruits" have specialized in creating n EW 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 bet Ween both of them.
A big topic of discussion inside the "How should the new creations be called?" A mixture between an apple and a pear could is called an apple-pear, the course, but this doesn ' t sound very interesting. The boss finally decides 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 are no shorter string that have the SAM E property.

A combination of a cranberry and a boysenberry would therefore be called a "boysecranberry" or a "Craboysenberry", for Exa Mple.

Your job is-to-write a program, computes such a shortest name for a combination of of both given fruits. Your algorithm should be efficient, otherwise it's unlikely that it'll execute in the alloted time for long fruit names .

Inputeach line of the input contains, strings that represent the names of the fruits, is should. All names has a maximum length of between and only consist of alphabetic characters. Input is terminated by end of file.
Outputfor each test case, output the shortest name of the resulting fruit on one line. If more than one shortest name was possible, any one is acceptable.
Sample Input

Apple Peachananas bananapear Peach

Sample Output

Appleachbananaspearch

This problem is mainly used in the LCS and backtracking methods ~ ~ ~ when backtracking in order to let them sequentially output ... Need to make a mark ~ ~ ~

Ac-code:

#include <cstdio> #include <cstring>using namespace std;int mark[105][105];char s1[105],s2[105];void Print (int a,int b) {if (a==0&&b==0) return, else if (a==0) {print (a,b-1);p rintf ("%c", S2[b-1]);} else if (b==0) {print (a-1,b);p rintf ("%c", S1[a-1]);} else if (mark[a][b]==1) {print (a-1,b-1);p rintf ("%c", S1[a-1]);} else if (mark[a][b]==2) {print (a-1,b);p rintf ("%c", S1[a-1]);} Else{print (a,b-1);p rintf ("%c", S2[b-1]);}} int main () {int Len1,len2,i,j;int dp[105][105];while (~scanf ("%s%s", &s1,&s2)) {memset (dp,0,sizeof (DP)); memset (Mark,0,sizeof (Mark)); Len1=strlen (S1); Len2=strlen (S2); for (i=1;i<=len1;i++) for (j=1;j<=len2;j++) {if ( S1[i-1]==s2[j-1]) {dp[i][j]=dp[i-1][j-1]+1;mark[i][j]=1;} else if (Dp[i-1][j]>dp[i][j-1]) {dp[i][j]=dp[i-1][j];mark[i][j]=2;} else{dp[i][j]=dp[i][j-1];mark[i][j]=3;}} Print (LEN1,LEN2);p rintf ("\ n");} return 0;}

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hdu1503:advanced Fruits "LCS"