POJ 1226 Substrings(KMP+枚举)

超级传送门：http://poj.org/problem?id=1226

题意要求一个最长的串X，其或其反转串rev(X)在所有串中均出现过，输出X的长度。

思路：先枚举第一个字符串的所有字串和字串的反转串，与剩下字符串进行串匹配，匹配过程中若二者有一个匹配成功则视为成功，从最长的字串开始枚举，第一个成功匹配的字串即为答案。

代码(注意reverse函数调用的位置，我之前多调用了一次导致WA)：

#include <cstdio>
#include <cstring>

using namespace std;

const int maxn = 105;
const int maxm = 105;

int P[maxm];
char A[maxn], B[maxm];
char S[105][maxn];

int n, m;

void preprocess(char* B, int* P, int m = -1)
{
    P[0] = -1;
    int j = -1;

    if (m == -1)
        m = strlen(B);

    for (int i = 1; i < m; i++)
    {
        while (j != -1 && B[j + 1] != B[i])
            j = P[j];

        if (B[j + 1] == B[i])
            j++;
        P[i] = j;
    }
}

int kmp(char* A, char* B, int* P, int n = -1, int m = -1)
{
    int j = -1;

    if (n == -1)
        n = strlen(A);
    if (m == -1)
        m = strlen(B);

    for (int i = 0; i < n; i++)
    {
        while (j != -1 && B[j + 1] != A[i])
            j = P[j];

        if (B[j + 1] == A[i])
            j++;
        if (j == m - 1)
            return 1;
    }
    return 0;
}

void reverse(char* S, int len = -1)
{
    if (m == -1)
        m = strlen(S);
    int halflen = len >> 1;

    for (int i = 0; i < halflen; i++)
    {
        char tmp = S[i];
        S[i] = S[len - i - 1];
        S[len - i - 1] = tmp;
    }
}

int main()
{
    int x, ansLen, t, ok;

    scanf("%d", &t);
    while (t--)
    {
        scanf("%d", &x);

        int len = 104;

        for (int i = 0; i < x; i++)
        {
            scanf("%s", S[i]);
            int tmpLen = strlen(S[i]);
            len = len < tmpLen ? len : tmpLen;
        }

        ansLen = 0;
        ok = 0;

        for (int i = len; i >= 1 && !ok; i--)
        {
            for (int j = 0; j <= len - i; j++)
            {
                char tmp[105];
                int tIndex = 0;
                for (int k = j; k < j + i; k++)
                    tmp[tIndex++] = S[0][k];
                tmp[tIndex] = '\0';
                strcpy(B, tmp);

                preprocess(B, P);

                int y;
                for (y = 1; y < x; y++)
                {
                    int retval = kmp(S[y], B, P);
                    reverse(B, i);
                    int revretval = kmp(S[y], B, P);
                    if (!retval && !revretval)
                        break;
                }

                if (y == x)
                {
                    ansLen = i;
                    ok = 1;
                    break;
                }

                preprocess(B, P);

                for (y = 1; y < x; y++)
                {
                    int retval = kmp(S[y], B, P);
                    reverse(B, i);
                    int revretval = kmp(S[y], B, P);
                    if (!retval && !revretval)
                        break;
                }

                if (y == x)
                {
                    ansLen = i;
                    ok = 1;
                    break;
                }
            }
        }

        printf("%d\n", ansLen);
    }

    return 0;
}