hdu 1238 Substrings(KMP)

题目链接：hdu 1238 Substrings

题目大意：给出若干个字符串，求一个最长的公共子串，该公共子串可以以逆序的形式存在在字符串中。

解题思路：枚举第一个串的开头做为匹配串，然后让该串和其他所有字符串求最长公共串的时候要去字符串正序和逆序中较大的。

#include <stdio.h>
#include <string.h>
#include <algorithm>

using namespace std;
const int N = 105;
int n, len[N], m, jump[N];
char s[N][N], f[N][N], t[N];

void init () {
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%s", s[i]+1);
		len[i] = strlen (s[i]+1);

		for (int j = 1; j <= len[i]; j++)
			f[i][j] = s[i][len[i]-j+1];
	}
}

void getJump () {

	int p = 0;
	for (int i = 2; i <= m; i++) {
		while (p > 0 && t[p+1] != t[i])
			p = jump[p];

		if (t[p+1] == t[i])
			p++;

		jump[i] = p;
	}
}

int KMP (int j) {
	int ans = 0, p = 0;
	for (int i = 1; i <= len[j]; i++) {
		while (p > 0 && t[p+1] != s[j][i])
			p = jump[p];

		if (t[p+1] == s[j][i])
			p++;

		ans = max(ans, p);
		if (p == m) p = jump[p];
	}

	p = 0;
	for (int i = 1; i <= len[j]; i++) {
		while (p > 0 && t[p+1] != f[j][i])
			p = jump[p];

		if (t[p+1] == f[j][i])
			p++;

		ans = max(ans, p);
		if (p == m) p = jump[p];
	}

	return ans;
}

int solve () {
	int ans = 0;

	for (int i = 1; i <= len[0]; i++) {
		strcpy (t+1, s[0]+i);
		m = strlen(t+1);

		int tmp = m;
		getJump ();
		for (int j = 1; j < n; j++)
			tmp = min (tmp, KMP (j));

		ans = max(ans, tmp);
	}
	return ans;
}

int main () {
	int cas;
	scanf("%d", &cas);
	while (cas--) {
		init ();
		printf("%d\n", solve ());
	}
	return 0;
}

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hdu 1238 Substrings(KMP)

原文：http://blog.csdn.net/keshuai19940722/article/details/21561001