HDU 1503 最长公共子序列的变形（重点在输出）

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

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
   
   
   
   
   
   
   
   


   
   
   
   

  

#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;
char a[1005],b[1005];
int dp[1005][1005];
int x[1005][1005];
void print( int m,int n)
{
    if(m==0&&n==0)
       return;
    else if(m==0&&n!=0)
    {
        print(m,n-1);
        printf("%c",b[n-1]);
    }
    else if(m!=0&&n==0)
    {
        print(m-1,n);
        printf("%c",a[m-1]);
    }
    else if(x[m][n]==0)
    {
        print(m-1,n-1);
        printf("%c",a[m-1]);
    }
    else if(x[m][n]==1)
    {
        print(m-1,n);
        printf("%c",a[m-1]);
    }
    else
    {
        print(m,n-1);
        printf("%c",b[n-1]);
    }
}
int main()
{
    while(~scanf("%s%s",a,b))
    {
        memset(dp,0,sizeof(dp));
        int n=strlen(a);
        int m=strlen(b);
        for(int i=1;i<=n;i++)
           for(int j=1;j<=m;j++)
           {
               if(a[i-1]==b[j-1])
               {
                   dp[i][j]=dp[i-1][j-1]+1;
                   x[i][j]=0;
               }
               else if(dp[i-1][j]>=dp[i][j-1])
               {
                   dp[i][j]=dp[i-1][j];
                   x[i][j]=1;
               }
               else if(dp[i-1][j]<dp[i][j-1])
               {
                   dp[i][j]=dp[i][j-1];
                   x[i][j]=2;
               }
           }
        print(n,m);
        printf("\n");
    }
    return 0;
}

方法二：非递归输出的方法

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
#define size 201
struct rem
{
    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*/
};

char a[size],b[size];
int dp[size][size];
rem ans[size];
int len;/*指示ans的长度*/
int main()
{
    int a1,b1,i,j,k;
    while (scanf("%s%s",a+1,b+1)!=EOF)
    {
        a1 = strlen(a+1);
        b1 = strlen(b+1);
        memset(dp,0,sizeof(dp));
        len = 0;
        for (i=1; i<=a1; i++)
        {
            for (j=1; j<=b1; j++)
            {
                if(a[i]==b[j])
                    dp[i][j] = dp[i-1][j-1]+1;
                else
                    dp[i][j] = max(dp[i-1][j],dp[i][j-1]);
            }
        }
        /*取出最长公共子序列的字母*/
        i = a1,j = b1;
        while (i!=0&&j!=0)
        {
            if ((dp[i][j] == dp[i-1][j-1]+1)&&a[i] == b[j])
            {
                ans[len].i = i;
                ans[len].j = j;
                ans[len++].ch = a[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--;
        }
        /*取出最长公共子序列的字母*/
        i = j = 1;
        for (k=len-1; k>=0; k--)
        {
            while (i!=ans[k].i)//此举是为了输出最长公共子序列之前和之中的非公共部分
            {
                printf("%c",a[i]);
                i++;
            }
            while (j!=ans[k].j)//作用同上
            {
                printf("%c",b[j]);
                j++;
            }
            printf("%c",ans[k].ch);//输出公共子序列的成员
            i++,j++;
        }
        printf("%s%s\n",a+ans[0].i+1,b+ans[0].j+1);//输出最长公共子序列最后一个元素之后的非公共部分
    }
    return 0;
}