LCS和输出 hdu 1503（Advanced Fruits）

Advanced Fruits

 

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

 题意：给两个单词，组合成一个，公共部分只输出1次。

题解：LCS 变形。输出注意用了回溯。在LCS时加上标记。

<span style="font-size:18px;">#include<cstdio>#include<cstring>#include<algorithm>using namespace std;char a[110];char b[110];int t;int c[110][110];//标记int dp[110][110];//前i个a与前j个b组成的LCS长度void lcsput(int i,int j){//回溯输出if (!i &&!j)return ;if (c[i][j]==0){//公共部分lcsput(i-1,j-1);//回溯printf ("%c",a[i-1]);}else if (c[i][j]==1){//a的特有元素lcsput(i-1,j);printf ("%c",a[i-1]);}else{//b的特有元素lcsput(i,j-1);printf ("%c",b[j-1]);}}int main(){while (scanf ("%s %s",a,b)!=EOF){int i,j;int len1=strlen(a);int len2=strlen(b);memset (c,0,sizeof(c));memset (dp,0,sizeof(dp));for (i=0;i<len1;i++)c[i][0]=1;for (i=0;i<len2;i++)c[0][i]=-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;c[i][j]=0;}else if (dp[i-1][j]>=dp[i][j-1]){c[i][j]=1;dp[i][j]=dp[i-1][j];}else{c[i][j]=-1;dp[i][j]=dp[i][j-1];}}}lcsput(len1,len2);printf ("\n");}return 0;} </span>

类似题： poj  Compromise   题解（大神）