当前位置：网站首页>hdu 1503 Advanced Fruits（最长公共子序列的应用）

hdu 1503 Advanced Fruits（最长公共子序列的应用）

2022-08-09 20:09:00 【51CTO】


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

Advanced Fruits


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2134    Accepted Submission(s): 1088
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
      
      

      
      
    
       
       
  • 1.
    • 
       
       
  • 2.
    • 
       
       
  • 3.
    •

Sample Output 

      
      

       
       appleach
       
       
bananas
       
       
pearch
      
      

      
      
    
       
       
  • 1.
    • 
       
       
  • 2.
    • 
       
       
  • 3.
    •

分析：大意，寻找包含两个字符串所有单个字符的最短字符串。


hdu 1503 Advanced Fruits（最长公共子序列的应用）_最长公共子序列


联想到最长公共子序列，输出那个寻找最长公共子序列的路径就是最小的最大串。


处理好边界，第0行全部指向左边，第0列全部指向上边，这样使得最终的汇聚点是（0,0），也就是递归输出的终止点。


      
      

       
       #include <iostream>
       
       

       
       #include <cstdio>
       
       

       
       #include <cstring>
       
       

       
       using 
       
       namespace 
       
       std;
       
       

       
       const 
       
       int 
       
       maxn
       
       =
       
       105;
       
       

       
       char 
       
       s1[
       
       maxn],
       
       s2[
       
       maxn];
       
       

       
       int 
       
       dp[
       
       maxn][
       
       maxn],
       
       pre[
       
       maxn][
       
       maxn];
       
       

       
       int 
       
       len1,
       
       len2;
       
       

       
       void 
       
       LCS(){
       
       
     
       
       memset(
       
       dp,
       
       0,
       
       sizeof(
       
       dp));
       
       
     
       
       memset(
       
       pre,
       
       0,
       
       sizeof(
       
       pre));
       
       
     
       
       len1
       
       =
       
       strlen(
       
       s1
       
       +
       
       1);
       
       
     
       
       len2
       
       =
       
       strlen(
       
       s2
       
       +
       
       1);
       
       
     
       
       for(
       
       int 
       
       i
       
       =
       
       1;
       
       i
       
       <=
       
       len1;
       
       i
       
       ++){
       
       
        
       
       for(
       
       int 
       
       j
       
       =
       
       1;
       
       j
       
       <=
       
       len2;
       
       j
       
       ++){
       
       
            
       
       if(
       
       s1[
       
       i]
       
       ==
       
       s2[
       
       j]){
       
       
                
       
       dp[
       
       i][
       
       j]
       
       =
       
       dp[
       
       i
       
       -
       
       1][
       
       j
       
       -
       
       1]
       
       +
       
       1;
       
       
                
       
       pre[
       
       i][
       
       j]
       
       =
       
       1; 
       
       //left:0; left-above:1; above:2;
       
       
            }
       
       
            
       
       else 
       
       if(
       
       dp[
       
       i
       
       -
       
       1][
       
       j]
       
       >
       
       dp[
       
       i][
       
       j
       
       -
       
       1]){
       
       
                
       
       dp[
       
       i][
       
       j]
       
       =
       
       dp[
       
       i
       
       -
       
       1][
       
       j];
       
       
                
       
       pre[
       
       i][
       
       j]
       
       =
       
       2;
       
       
            }
       
       
            
       
       else {
       
       
                
       
       dp[
       
       i][
       
       j]
       
       =
       
       dp[
       
       i][
       
       j
       
       -
       
       1];
       
       
                
       
       pre[
       
       i][
       
       j]
       
       =
       
       0;
       
       
            }
       
       
        }
       
       
     }
       
       
}
       
       

       
       void 
       
       print(
       
       int 
       
       x,
       
       int 
       
       y){
       
       
    
       
       if(
       
       x
       
       ==
       
       0
       
       &&
       
       y
       
       ==
       
       0) 
       
       return ;
       
       
    
       
       if(
       
       pre[
       
       x][
       
       y]
       
       ==
       
       0) {  
       
       print(
       
       x,
       
       y
       
       -
       
       1); 
       
       printf(
       
       "%c",
       
       s2[
       
       y]); }
       
       
    
       
       else 
       
       if(
       
       pre[
       
       x][
       
       y]
       
       ==
       
       1){ 
       
       print(
       
       x
       
       -
       
       1,
       
       y
       
       -
       
       1); 
       
       printf(
       
       "%c",
       
       s1[
       
       x]); }
       
       
    
       
       else {  
       
       print(
       
       x
       
       -
       
       1,
       
       y); 
       
       printf(
       
       "%c",
       
       s1[
       
       x]);  }
       
       
}
       
       

       
       int 
       
       main()
       
       
{
       
       
    
       
       //freopen("cin.txt","r",stdin);
       
       
    
       
       while(
       
       ~scanf(
       
       "%s%s",
       
       s1
       
       +
       
       1,
       
       s2
       
       +
       
       1)){
       
       
        
       
       LCS();
       
       
        
       
       for(
       
       int 
       
       i
       
       =
       
       1;
       
       i
       
       <=
       
       len2;
       
       i
       
       ++) 
       
       pre[
       
       0][
       
       i]
       
       =
       
       0; 
       
       //边界处理很重要
       
       
        
       
       for(
       
       int 
       
       i
       
       =
       
       1;
       
       i
       
       <=
       
       len1;
       
       i
       
       ++) 
       
       pre[
       
       i][
       
       0]
       
       =
       
       2;
       
       
        
       
       int 
       
       i
       
       =
       
       len1,
       
       j
       
       =
       
       len2,
       
       top
       
       =
       
       0;
       
       
        
       
       char 
       
       ans[
       
       maxn];
       
       
        
       
       print(
       
       len1,
       
       len2);
       
       
        
       
       puts(
       
       "");
       
       
    }
       
       
    
       
       return 
       
       0;
       
       
}
      
      

      
      
    
       
       
  • 1.
    • 
       
       
  • 2.
    • 
       
       
  • 3.
    • 
       
       
  • 4.
    • 
       
       
  • 5.
    • 
       
       
  • 6.
    • 
       
       
  • 7.
    • 
       
       
  • 8.
    • 
       
       
  • 9.
    • 
       
       
  • 10.
    • 
       
       
  • 11.
    • 
       
       
  • 12.
    • 
       
       
  • 13.
    • 
       
       
  • 14.
    • 
       
       
  • 15.
    • 
       
       
  • 16.
    • 
       
       
  • 17.
    • 
       
       
  • 18.
    • 
       
       
  • 19.
    • 
       
       
  • 20.
    • 
       
       
  • 21.
    • 
       
       
  • 22.
    • 
       
       
  • 23.
    • 
       
       
  • 24.
    • 
       
       
  • 25.
    • 
       
       
  • 26.
    • 
       
       
  • 27.
    • 
       
       
  • 28.
    • 
       
       
  • 29.
    • 
       
       
  • 30.
    • 
       
       
  • 31.
    • 
       
       
  • 32.
    • 
       
       
  • 33.
    • 
       
       
  • 34.
    • 
       
       
  • 35.
    • 
       
       
  • 36.
    • 
       
       
  • 37.
    • 
       
       
  • 38.
    • 
       
       
  • 39.
    • 
       
       
  • 40.
    • 
       
       
  • 41.
    • 
       
       
  • 42.
    • 
       
       
  • 43.
    • 
       
       
  • 44.
    • 
       
       
  • 45.
    • 
       
       
  • 46.
    • 
       
       
  • 47.
    • 
       
       
  • 48.
    • 
       
       
  • 49.
    • 
       
       
  • 50.
    •



原网站

版权声明
本文为[51CTO]所创，转载请带上原文链接，感谢
https://blog.51cto.com/u_15746559/5560973

边栏推荐

猜你喜欢

随机推荐