标签：existing service therefore getch ade tmp name ora integer
InputThe 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.OutputFor 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
Sample Output
abb IDENTITY LOST
#include<iostream> #include<algorithm> #include<cstdio> #include<vector> #include <sstream> #include<string> #include<cstring> #include<list> using namespace std; #define MAXN 202 #define INF 0x3f3f3f3f #define N 4002 typedef long long LL; /* 枚举第一个串的所有子串 找到就停止 */ char s[N][MAXN]; int Next[MAXN]; void kmp_pre(char t[]) { int m = strlen(t); int j,k; j = 0;k = Next[0] = -1; while(j<m) { if(k==-1||t[j]==t[k]) Next[++j] = ++k; else k = Next[k]; } } bool KMP(char t[],char s[]) { int n = strlen(s),m=strlen(t); int i,j; j = 0; for(i=0;i<n;i++) { while(j>0&&s[i]!=t[j]) j = Next[j]; if(s[i]==t[j]) j++; if(j>=m) return true; } return false; } int main() { int n; while(scanf("%d",&n),n) { int len = INF,k=-1; for(int i=0;i<n;i++) { scanf("%s",s[i]); if(len>strlen(s[i])) { len = strlen(s[i]); k = i; } } bool f = false; char ans[MAXN]; for(int l=len;l>0;l--) { for(int i=0;i+l<=len;i++) { char t[MAXN] = {0}; strncpy(t,s[k]+i,l); kmp_pre(t); int j; //cout<<t<<endl; for(j=0;j<n;j++) { if(j==k) continue; if(KMP(t,s[j])) continue; else break; } if(j==n) { if(!f) { strcpy(ans,t); f = true; } else { if(strcmp(ans,t)>0) strcpy(ans,t); } } if(f&&(i+l==len)) { printf("%s\n",ans); i = INF; l = -1; break; } } } if(!f) printf("IDENTITY LOST\n"); } }
只需要枚举所有后缀，然后在其他串中找最长相同前缀即可
#include <stdio.h> #include <string.h> const int MAX_N = 4001; const int MAX_CHS = 201; const int ARR_SIZE = 26; int N; char res[MAX_CHS], dict[MAX_N][MAX_CHS]; short next[MAX_CHS]; inline int min(int a, int b) { return a < b ? a : b; } inline int max(int a, int b) { return a > b ? a : b; } void getNext(char *chs, int len) { memset(next, 0, sizeof(short)*len); for (int i = 1, j = 0; i < len; ) { if (chs[i] == chs[j]) next[i++] = ++j; else if (j > 0) j = next[j-1]; else i++; } } int getLogestPre(char *chs, int len) { getNext(chs, len); for (int i = 1; i < N; i++) { char *p = dict[i]; int j = 0, tmp = 0; for (; *p && j < len; ) { if (*p == chs[j]) { p++, j++; tmp = max(tmp, j); } else if (j > 0) j = next[j-1]; else p++; } len = tmp; } return len; } int main() { while (scanf("%d", &N) && N) { getchar(); // don‘t forget to get rid of ‘\‘ for (int i = 0; i < N; i++) { gets(dict[i]); } int len = strlen(dict[0]), ans = 0, pos = 0; for (int i = 0; i < len; i++) { int tmp = getLogestPre(dict[0]+i, len-i); if (tmp >= ans) { if (tmp > ans) { ans = tmp; pos = i; } else { bool smaller = true; for (int t = 0; t < ans; t++) { if (dict[0][pos+t] > dict[0][i+t]) break; else if (dict[0][pos+t] < dict[0][i+t]) { smaller = false; break; } } if (smaller) pos = i; } } } if (ans) { for (int i = 0; i < ans; i++) { putchar(dict[0][pos+i]); } putchar(‘\n‘); } else puts("IDENTITY LOST"); } return 0; }
标签：existing service therefore getch ade tmp name ora integer
原文地址：http://www.cnblogs.com/joeylee97/p/6681758.html
