HDU 1238 Substrings kmp

一、內容

 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 file 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

二、思路

– 暴力枚舉子串即可。 注意子串還需要左右顛倒比較一次。

三、代碼

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 105;
char s[N][N], p[N];
int t, n, m, k, mx, ne[N];
void getNext() {
	ne[1] = 0;
	for (int i = 2, j = 0; i <= m; i++) {
		while (j && p[i] != p[j + 1]) j = ne[j];
		if (p[i] != p[j + 1]) j++;
		ne[i]= j;
	}
}
bool kmp(int k) {
	n = strlen(s[k] + 1);
	getNext();
	for (int i = 1, j = 0; i <= n; i++) {
		while (j && s[k][i] != p[j + 1]) j = ne[j];
		if (s[k][i] == p[j + 1]) j++;
		if (j == m) return true; 
	} 
 	//左右顛倒 
 	reverse(p + 1, p + 1 + m);
 	getNext();
	for (int i = 1, j = 0; i <= n; i++) {
		while (j && s[k][i] != p[j + 1]) j = ne[j];
		if (s[k][i] == p[j + 1]) j++;
		if (j == m) return true; 
	} 
 	return false;
} 
bool ok() {
	for (int i = 1; i <= k; i++) {
		if (!kmp(i)) return false;
	}
	return true;
}
int main() {
	scanf("%d", &t);
	while (t--) {
		scanf("%d", &k);
		for (int i = 1; i <= k; i++) scanf("%s", s[i] + 1);
		n = strlen(s[1] + 1); mx = 0;
		for (int i = 1; i <= n; i++) {
			for (int j = i; j <= n; j++) {
				m = j - i + 1;
				for (int h = i; h <= j; h++) p[h - i + 1] = s[1][h];
				p[m + 1] = '\0';
				if (ok() && m > mx) mx = m;
			}		
		}
		printf("%d\n", mx);
	} 
	return 0; 
}