HDU - 1503 Advanced Fruits

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

首先要看懂题意，题目说两个亲本的名字要作为新一代植株名字的母串，且新名字要尽可能的短。

思路很明显，最长公共字符串问题，两个名字中相同的字符串肯定只需要输出一次就够了，然后将剩下的字符按顺序输出即可。

当时不知道怎么按顺序输出，一直在考虑两个字符串同步输出，后来才想到先定下一个，然后在需要时再输出另一个的方法。
两个字符串同时输出，肯定是不可能的啊！竟然犯这么低级的错误，太恶心了。。。。

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<stack>
#include<deque>
#include<iostream>
using namespace std;
typedef long long  LL;
char str1[110],str2[110];
int dp[110][110];
void output(int i,int j)
{
    if(i==0||j==0)
        return ;
    else
    {
        if(str1[i-1]==str2[j-1])
        {
            output(i-1,j-1);
            str1[i-1]-=90;
            str2[j-1]-=90;
        }
        else
        {
            if(dp[i][j-1]>=dp[i-1][j])
                output(i,j-1);
            else
                output(i-1,j);

        }
    }
}
int main()
{
    int i,p,j,k;
    int len1,len2;
    while(scanf("%s%s",str1,str2)!=EOF)
    {
        getchar();
        memset(dp,0,sizeof(dp));
        len1=strlen(str1);
        len2=strlen(str2);
        for(i=1; i<=len1; i++)
            for(j=1; j<=len2; j++)
            {
                if(str1[i-1]==str2[j-1])
                    dp[i][j]=dp[i-1][j-1]+1;
                else
                    dp[i][j]=max(dp[i][j-1],dp[i-1][j]);
            }
        k=0;
        output(len1,len2);
        /*
        for(i=0;i<len1;i++)
            if(str1[i]<50)
                printf("%c",str1[i]+90);*/

        for(i=0;i<len1;i++)
        {

            if(str1[i]<50)
            {
                char mid=str1[i]+90;
                for(j=k;j<len2;j++)
                {
                    if(str2[j]<50)
                    {
                        str2[j]=0;
                        printf("%c",mid);
                        break;
                    }
                    else
                    {
                        printf("%c",str2[j]);
                    }
                }
                k=j+1;
            }
            else
                printf("%c",str1[i]);
        }
        for(j=k;j<len2;j++)
            printf("%c",str2[j]);
        putchar('\n');
    }
    return 0;
}