hdu 1503:Advanced Fruits（动态规划 DP & 最长公共子序列（LCS）问题升级版）

Advanced Fruits

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1360    Accepted Submission(s): 672
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

 

 

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　　  这是第二次坑在这道题上了，试验了好几种思路不就是运行时错误就是WA，最后火了还是看了别人的博客才做出来。

　　思路很简单，在LCS的基础之上加上路径记录。生成dp数组的时候做上标记，之后按顺序输出结果字符串。

　　我坑就坑在我没有想到好的记录路径的方法，还是高手的方法精炼，准确又简单。

　　下面是代码，注意数组的初始化。

 

 1 #include <iostream>
 2 #include <string.h>
 3 using namespace std;
 4 int dp[1001][1001];
 5 int f[1001][1001];
 6 char a[1000],b[1000];
 7 int l1,l2;
 8 int Length(char a[])    //返回一字符串长度 
 9 {
10     int i;
11     for(i=0;a[i]!='\0';i++);
12     return i;
13 }
14 void lcs_pre(char a[],char b[])    //进行标记 
15 {
16     l1 = Length(a);
17     l2 = Length(b);
18     for(int i=0;i<=l1;i++){
19         dp[i][0] = 0;
20         f[i][0] = 3;
21     }
22     for(int i=0;i<=l2;i++){
23         dp[0][i] = 0;
24         f[0][i] = 1;
25     }
26     for(int i=1;i<=l1;i++)
27         for(int j=1;j<=l2;j++){
28             if( a[i-1]==b[j-1] ){
29                 dp[i][j] = dp[i-1][j-1] + 1;
30                 f[i][j] = 2;
31             }
32             else if( dp[i][j-1]>=dp[i-1][j] ){
33                 dp[i][j] = dp[i][j-1];
34                 f[i][j] = 1; 
35             }
36             else{
37                 dp[i][j] = dp[i-1][j];
38                 f[i][j] = 3;
39             }
40         }
41 }
42 void Print(int x,int y)    //输出结果字符串 
43 {
44     if(x==0 && y==0)
45         return ;
46     switch(f[x][y]){
47         case 1:
48             Print(x,y-1);
49             cout<<b[y-1];
50             break;
51         case 2:
52             Print(x-1,y-1);
53             cout<<a[x-1];
54             break;
55         case 3:
56             Print(x-1,y);
57             cout<<a[x-1];
58             break;
59         default:break;
60     }
61     return ;
62 }
63 int main()
64 {
65     while(cin>>a>>b){ 
66         l1 = Length(a);
67         l2 = Length(b);
68         lcs_pre(a,b);    //进行标记
69         Print(l1,l2);
70         cout<<endl;
71     }
72     return 0;
73 }

 

Freecode : www.cnblogs.com/yym2013