hdu 1503 Advanced Fruits

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1503

题目描述：

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

题目大意：

给你两个字符串，然后输出一个新的字符串，要求两个字符串的最长公共字串只输出一次

题目分析：

同求最长公共子序列，这里需要注意的地方就是这里需要输出，那么我们就开两个数组记录一下公共部分在两个字符串中的下标

Ac代码：

#include<iostream>#include<cstring>#include<cmath>#include<cstdio>#include<algorithm>using namespace std;int dp[105][105];char s1[105],s2[105];int cnt;struct node{    int x,y;//x记录s1的起始位置，y记录s2的起始位置    char tmp;//tmp记录字符}v[105];void LCS(int m,int n){    memset(dp,0,sizeof(dp));    for(int i=1;i<=m;i++)    {        for(int j=1;j<=n;j++)        {            if(s1[i]==s2[j])                dp[i][j]=dp[i-1][j-1]+1;            else                dp[i][j]=max(dp[i-1][j],dp[i][j-1]);        }    }//dp[i][j]代表s1从1->i，s2从1->j最长公共子序列的长度    cnt=1;    int len=dp[m][n];    if(!dp[m][n])        return;    else    {        int i=m,j=n;        while(len>0)        {            if(s1[i]==s2[j])            {                v[cnt].x=i;//记录起始位置                v[cnt].y=j;                v[cnt].tmp=s1[i];                i--;                j--;                len--;                cnt++;            }            else if(dp[i-1][j]>dp[i][j-1])                i--;            else                j--;        }    }}int main(){    while(scanf("%s%s",s1+1,s2+1)!=EOF)    {        int m=strlen(s1+1),n=strlen(s2+1);        LCS(m,n);        cnt--;        int i=1,j=1;        while(i<=m||j<=n)        {            while(i!=v[cnt].x&&i<=m)            {                printf("%c",s1[i]);                i++;            }            while(j!=v[cnt].y&&j<=n)            {                printf("%c",s2[j]);                j++;            }            if(cnt>=1)            {                printf("%c",v[cnt].tmp);                cnt--;                i++;                j++;            }        }        printf("\n");    }    return 0;}