HDU-1503 Advanced Fruits（LCS）

Advanced Fruits

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 4069 Accepted Submission(s): 2144

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 peach

ananas banana

pear peach

Sample Output

appleach

bananas

pearch

Source

University of Ulm Local Contest 1999

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=300;

char str1[maxn],str2[maxn];
int dp[maxn][maxn],len; //len指示ans的长度

struct node{
int i,j;    //i记录主串位置，j记录副串当前字符位置
char ch;        //记录当前字符
}ans[maxn];

int max(int a,int b){
return a>b?a:b;
}

void LCS(int m,int n){
memset(dp,0,sizeof(dp));
int i,j;
for(i=1;i<=m;i++)//正常的LCS查找相同字符的长度
for(j=1;j<=n;j++)
if(str1[i]==str2[j])
dp[i][j]=dp[i-1][j-1]+1;
else
dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
if(dp[m]
==0){    //如果没有公共序列，直接输出
printf("%s%s",str1,str2);
}
else{
i=m;j=n;
len=0;
while(i!=0 && j!=0){    //取出最长公共子序列的字母
if((dp[i][j]==dp[i-1][j-1]+1) && str1[i]==str2[j]){
ans[len].i=i;
ans[len].j=j;
ans[len++].ch=str1[i];      //倒序保存最长公共子序列字母
i--;j--;
}
else if(dp[i-1][j]>dp[i][j-1])
i--;
else
j--;
}
}
}

int main(){
int len1,len2,i,j,k;
while(scanf("%s%s",str1+1,str2+1)!=EOF){
len1=strlen(str1+1);
len2=strlen(str2+1);
LCS(len1,len2);
i=j=1;
for(k=len-1;k>=0;k--){//从左往右两个字符串依次输出
while(i!=ans[k].i){//如果不是重合的字符，输出字串1的字符
printf("%c",str1[i]);
i++;
}
while(j!=ans[k].j){//如果不是重合的字符，输出字串2的字符
printf("%c",str2[j]);
j++;
}
printf("%c",ans[k].ch);//输出两个字串重合的字符
i++;j++;
}
printf("%s%s\n",str1+1+ans[0].i,str2+1+ans[0].j);//输出两个字串中最右个重合字符右边，剩下得到字符。
}
return 0;
}