HDOJ--1503 Advanced Fruits

Advanced Fruits

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2025    Accepted Submission(s): 1035
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

appleachbananas
pearch
题目大意：额，这道题的大概意思是两个单词拼接成同一个单词，拼接的方式就是把共有的最长子序列抵消掉，然后输出就行了。
详见代码。
ac代码：
#include<stdio.h>#include<string.h>#define max(a,b) a>b?a:bint dp[1010][1010];struct node{int x,y;char c;}rp[1010];int main(){char a[1010],b[1010];while(scanf("%s %s",&a,&b)!=EOF){ //这个循环也是基本的模板，就是DP求最长子序列。 int i,j,k;int len1=strlen(a);int len2=strlen(b);memset(dp,0,sizeof(dp));for(i=1,k=-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;}elsedp[i][j]=max(dp[i-1][j],dp[i][j-1]);}if(dp[len1][len2]==0)printf("%s%s",a,b);else{i=len1,j=len2,k=-1;while(i!=0&&j!=0){//这个while循环用来记录子序列在两个原始序列中的位置。 if((dp[i][j]==dp[i-1][j-1]+1)&&a[i-1]==b[j-1]){rp[++k].x=i-1;rp[k].y=j-1;rp[k].c=a[i-1];i--;j--;}else if(dp[i-1][j]>dp[i][j-1])i--;elsej--;}int key=k;i=j=0;for(k=key;k>=0;k--){//这一部分是输出，分三部分输出，第一部分 不是子序列但是在最后一个子序列前面的字符。 while(i!=rp[k].x){printf("%c",a[i]);//第二部分，就是子序列这一部分的字符。 i++;}while(j!=rp[k].y){printf("%c",b[j]);j++;}printf("%c",rp[k].c);i++;j++;}printf("%s%s\n",a+rp[0].x+1,b+rp[0].y+1);//第三部分就是子序列之后的字符。 }}return 0;}