hdu1503 Advanced Fruits
2016-05-20 20:38
435 查看
思路:LCS模板题...需要记录路径
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
Sample Output
#include <stdio.h> #include <string.h> #include <algorithm> using namespace std; char s1[1000],s2[1000]; int len1,len2,dp[1000][1000],mark[1000][1000]; void LCS() { int i,j; memset(dp,0,sizeof(dp)); for(i = 0;i<=len1;i++) mark[i][0] = 1; for(i = 0;i<=len2;i++) mark[0][i] = -1; for(i = 1; i<=len1; i++) { for(j = 1; j<=len2; j++) { if(s1[i-1]==s2[j-1]) { dp[i][j] = dp[i-1][j-1]+1; mark[i][j] = 0; } else if(dp[i-1][j]>=dp[i][j-1]) { dp[i][j] = dp[i-1][j]; mark[i][j] = 1; } else { dp[i][j] = dp[i][j-1]; mark[i][j] = -1; } } } } void PrintLCS(int i,int j) { if(!i && !j) return ; if(mark[i][j]==0) { PrintLCS(i-1,j-1); printf("%c",s1[i-1]); } else if(mark[i][j]==1)//根据回溯的位置进行输出 { PrintLCS(i-1,j); printf("%c",s1[i-1]); } else { PrintLCS(i,j-1); printf("%c",s2[j-1]); } } int main() { while(~scanf("%s%s",s1,s2)) { len1 = strlen(s1); len2 = strlen(s2); LCS(); PrintLCS(len1,len2); printf("\n"); } return 0; }
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
相关文章推荐
- TensorFlow实战之:Quick Start
- 如何监听Tabbar的点击
- A. Infinite Sequence
- UITextField && UITextView
- UIMenuController
- MySQL入门--字段的唯一约束UNIQUE
- Codeforces Round #350 (Div. 2) E. Correct Bracket Sequence Editor【模拟+链表】
- lightoj 1421 - Wavio Sequence LIS变形
- querySelectorAll对象无ID的题
- iOS-Network学习笔记(一)——NSURLRequest/NSURLResponse
- ue4局域网加入流程
- 一次关于更改UI导致autolayout系统崩溃的问题
- NGUI poplist使用
- 深入源码解析Android中的Handler,Message,MessageQueue,Looper
- Android UI库书签
- Android CPU使用率:top和dump cpuinfo的不同
- MQ通道配置示例(2):Server/Requester
- Android Volley 之自定义Request
- UGUI ScrollRect完美使用
- ligerui联动清空控件值