【题解】HDU-1238 Substrings

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HDU-1238 Substrings 类型 字符串 KMP

Substrings(HDU-1238)

题面

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.

输入

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.

输出

There should be one line per test case containing the length of the largest string found.

样例输入

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2
3
ABCD
BCDFF
BRCD
2
rose
orchid

样例输出

1
2
2
2

提示

无

思路

枚举子串，暴力KMP即可。

代码

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char s[105][mxn], t[mxn], t2[mxn];
int nxt[mxn], nxt2[mxn], len[mxn];

void getnxt(char* t, int m)
{
    int i = 0, j = -1; nxt[0] = -1;
    while (i < m)
    {
        if (j == -1 || t[i] == t[j]) {
            i++, j++;
            // if (t[i] == t[j])
            //     nxt[i] = nxt[j]; // next数组优化
            // else
                nxt[i] = j;
        } else
            j = nxt[j];
    }
}

int KMP(char* s, char* t, int n, int m)
{
    int i = 0, j = 0, ans = 0;
    while (i < n)
    {
        if (j == -1 || s[i] == t[j]) {
            i++, j++;
            if (j >= m) {   // 匹配
                // ans++;
                // j = nxt[j];
                return i-j;
            }
        } else
            j = nxt[j];
    }
    // return ans;
    return -1;
}

int main()
{
    int T; scanf("%d", &T);
    while(T--)
    {
        int n; scanf("%d", &n);
        for(int i=0; i<n; i++){
            scanf("%s", s[i]);
            len[i] = strlen(s[i]);
        }
        int ans=0;
        for(int i=0; i<len[0]; i++) // 枚举模式串起点
        {
            int tl = 0, k;
            for(int j=i; j<len[0]; j++) // 枚举模式串长度
            {
                t[tl++] = s[0][j];
                for(int x=0; x<tl; x++) t2[x] = t[tl-x-1];
                t[tl] = t2[tl] = '\0';
                getnxt(t, tl);
                getnxt(t2, tl);
                for(k=0; k<n; k++)  // 枚举所有文本串
                    if(KMP(s[k], t, len[k], tl) == -1 && KMP(s[k], t2, len[k], tl) == -1)
                        break;
                if(k>=n)    // 满足条件
                    ans = max(ans, tl);
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}