HSU 2328 Corporate Identity
2016-02-23 14:23
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Corporate Identity
Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 817 Accepted Submission(s): 325
[align=left]Problem Description[/align]
Beside other services, ACM helps companies to clearly state their “corporate identity”, which includes company logo but also other signs, like trademarks. One of such companies is Internet Building Masters (IBM), which has recently asked ACM for a help with their new identity. IBM do not want to change their existing logos and trademarks completely, because their customers are used to the old ones. Therefore, ACM will only change existing trademarks instead of creating new ones.
After several other proposals, it was decided to take all existing trademarks and find the longest common sequence of letters that is contained in all of them. This sequence will be graphically emphasized to form a new logo. Then, the old trademarks may still be used while showing the new identity.
Your task is to find such a sequence.
[align=left]Input[/align]
The input contains several tasks. Each task begins with a line containing a positive integer N, the number of trademarks (2 ≤ N ≤ 4000). The number is followed by N lines, each containing one trademark. Trademarks will be composed only from lowercase letters, the length of each trademark will be at least 1 and at most 200 characters.
After the last trademark, the next task begins. The last task is followed by a line containing zero.
[align=left]Output[/align]
For each task, output a single line containing the longest string contained as a substring in all trademarks. If there are several strings of the same length, print the one that is lexicographically smallest. If there is no such non-empty string, output the words “IDENTITY LOST” instead.
Sample Input
3 aabbaabb abbababb bbbbbabb 2 xyz abc 0
[align=left]Sample Output[/align]
abb IDENTITY LOST
题意是给定4000以内个字符串,找其中最长公共子串,但是要求如果有多个长度相同的话输出按字典序取最小。
这个跟HDU1238类似,不过这个可以考虑用KMP因为是不用考虑反向字符串的。
试了一下,还是直接用find比较快,代码如下:
/************************************************************************* > File Name: Corporate_Identity.cpp > Author: Zhanghaoran > Mail: chilumanxi@xiyoulinux.org > Created Time: 2016年02月23日 星期二 14时01分07秒 ************************************************************************/ #include <iostream> #include <algorithm> #include <cstring> #include <cstdio> #include <cstdlib> #include <string> using namespace std; int n; string str[4010]; int main(void){ while(1){ cin >> n; if(!n) break; int len = 110; int len_pos = 0; for(int i = 0; i < n; i ++){ cin >> str[i]; if(len > str[i].size()){ len = str[i].size(); len_pos = i; } } bool flag= false; string ans; bool vis = true; for(int i = len; i >= 1; i --){ for(int j = 0; i + j <= len; j ++){ string testa = str[len_pos].substr(j, i); for(int k = 0; k < n; k ++){ if(k == len_pos) continue; if(str[k].find(testa) == -1){ flag = true; break; } } if(!flag){ if(vis){ ans = testa; vis = false; } else{ if(testa < ans) ans = testa; } } flag = false; } if(!vis) break; } if(vis) cout << "IDENTITY LOST" << endl; else cout << ans << endl; } }
查看原文:http://chilumanxi.org/2016/02/23/hsu-2328-corporate-identity/
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