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Substrings
Description
You are given a number of case-sensitive strings of alphabetic characters, find the largest string X, such that either X, or its inverse can be found as a substring of any of the given strings.
Input
The first line of the input contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case contains a single integer n (1 <= n <= 100), the number of given strings, followed
by n lines, each representing one string of minimum length 1 and maximum length 100. There is no extra white space before and after a string.
Output
There should be one line per test case containing the length of the largest string found.
Sample Input 2 3 ABCD BCDFF BRCD 2 rose orchid Sample Output 2 2 Source |
code:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
#define rep(i, l, r) for (int i = l; i <= r; i++)
#define REP(i, l, r) for (int i = l; i >= r; i--)
#define MAXN 1010
int len, next[MAXN], a[MAXN][MAXN], T_T, n, minlen, l[MAXN], f[MAXN], ans;
inline void getnext(int *a, int len) {
memset(next, 0, sizeof(next));
next[0] = -1;
int i = 0, j = -1;
while(i < len-1) {
if (!~j || a[i] == a[j]) i++, j++, next[i] = j;
else j = next[j];
}
}
inline bool kmp(int *s, int n, int *a, int m) {
getnext(a, m);
int i = 0, j = 0;
while (i < n) {
if (!~j || s[i] == a[j]) i++, j++;
else j = next[j];
if (j == m) return 1;
}
return 0;
}
int main() {
cin >> T_T;
while (T_T--) {
ans = 0;
cin >> n;
minlen = 12345678;
char ch[MAXN];
int kk;
rep(i, 1, n) {
scanf("%s", ch);
if (int(strlen(ch)) < minlen) {
minlen = int(strlen(ch));
kk = 2*i - 1;
}
minlen = min(minlen, int(strlen(ch)));
rep(j, 0, strlen(ch)-1) a[2*i-1][j] = int(ch[j]);
rep(j, 0, strlen(ch)-1) a[2*i][strlen(ch)-1-j] = int(ch[j]);
l[2*i-1] = l[2*i] = strlen(ch);
}
rep(L, 0, minlen-1) {
rep(R, L, minlen-1) {
memset(f, 0, sizeof(f));
rep(i, 0, R-L) f[i] = a[kk][i+L];
bool flag = 0;
rep(i, 1, n)
if (!kmp(a[2*i-1], l[2*i-1], f, R-L+1) && !kmp(a[2*i], l[2*i], f, R-L+1)) {
flag = 1;
break;
}
if (!flag) ans = max(ans, R-L+1);
}
}
cout << ans << endl;
}
}kyeremal-poj1226-Substrings-模式匹配
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原文地址：http://blog.csdn.net/kyeremal/article/details/46293943
