POJ 3450 Corporate Identity
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Memory Limit: 65536KTotal Submissions: 4942
Accepted: 1858
Description
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.
Input
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.
Output
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
3aabbaabbabbababbbbbbbabb2xyzabc0
Sample Output
abbIDENTITY LOST
Source
题目链接 :http://poj.org/problem?id=3450
题目大意 :给出若干组主串,求这些主串的最长公共连续子串
题目分析 :从第一个主串开始枚举其后缀子串然后分别与其余主串用KMP算法进行匹配,详细解释见程序
#include <cstdio>#include <cstring>char text[4001][205]; //主串char subtext[205]; //模式串char res[205]; //答案串int next[205]; int T; //主串个数int sublen; //模式串长int max; //答案串串长的最大值void get_next() //模版得到next数组{ int i = 0, j = -1; next[0] = -1; while(subtext[i] != '\0') { if(j == -1 || subtext[i] == subtext[j]) { i++; j++; if(subtext[i] == subtext[j]) next[i] = next[j]; else next[i] = j; } else j = next[j]; }}void KMP() { memset(next,0,sizeof(next)); //每次调用,初始化next数组 get_next(); //获取next数组 int cur, j, k; max = 10000; for(int i = 1; i < T; i++) //从第二个主串开始枚举 { j = 0, k = 0, cur = 0; while(j < strlen(text[i]) && k < sublen) //KMP算法 { if(k == -1 || text[i][j] == subtext[k]) { j++; k++; } else k = next[k]; if(k > cur) //找到最长匹配长度 cur = k; } //要找从第二个开始的每个主串的最长匹配长度的最小值 //才能保证答案串与每个主串都能匹配 if(max > cur) max = cur; }}int main(){ int ans; //答案串串长实际值 while(scanf("%d",&T) != EOF && T) { memset(text,0,sizeof(text)); //初始化主串 ans = 0; for(int i = 0; i < T; i++) scanf("%s", text[i]); for(int i = 0; i <= 200; i++) //枚举第一个主串的每一个后缀 { sublen = 200 - i; strcpy(subtext, text[0] + i); //将其复制给subtext KMP(); //执行KMP算法 if(ans < max) { ans = max; //复制得到答案串,text[0]+i是第一个主串的第i个后缀 //复制长度为答案串串长 strncpy(res,text[0] + i,ans); res[ans] = '\0'; } //若答案串串长相等则输出字典序小的一个 else if(ans == max) { char temp[205]; strncpy(temp,text[0] + i, ans); temp[ans] = '\0'; if(strcmp(res,temp) > 0) strcpy(res,temp); } } if(ans == 0) printf("IDENTITY LOST\n"); else printf("%s\n",res); } return 0;}
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