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

12
23
3ABCD
4BCDFF
5BRCD
62
7rose
8orchid

样例输出

12
22

提示

无

思路

枚举子串，暴力KMP即可。

代码

 1char s[105][mxn], t[mxn], t2[mxn];
 2int nxt[mxn], nxt2[mxn], len[mxn];
 3
 4void getnxt(char* t, int m)
 5{
 6    int i = 0, j = -1; nxt[0] = -1;
 7    while (i < m)
 8    {
 9        if (j == -1 || t[i] == t[j]) {
10            i++, j++;
11            // if (t[i] == t[j])
12            //     nxt[i] = nxt[j]; // next数组优化
13            // else
14                nxt[i] = j;
15        } else
16            j = nxt[j];
17    }
18}
19
20int KMP(char* s, char* t, int n, int m)
21{
22    int i = 0, j = 0, ans = 0;
23    while (i < n)
24    {
25        if (j == -1 || s[i] == t[j]) {
26            i++, j++;
27            if (j >= m) {   // 匹配
28                // ans++;
29                // j = nxt[j];
30                return i-j;
31            }
32        } else
33            j = nxt[j];
34    }
35    // return ans;
36    return -1;
37}
38
39int main()
40{
41    int T; scanf("%d", &T);
42    while(T--)
43    {
44        int n; scanf("%d", &n);
45        for(int i=0; i<n; i++){
46            scanf("%s", s[i]);
47            len[i] = strlen(s[i]);
48        }
49        int ans=0;
50        for(int i=0; i<len[0]; i++) // 枚举模式串起点
51        {
52            int tl = 0, k;
53            for(int j=i; j<len[0]; j++) // 枚举模式串长度
54            {
55                t[tl++] = s[0][j];
56                for(int x=0; x<tl; x++) t2[x] = t[tl-x-1];
57                t[tl] = t2[tl] = '\0';
58                getnxt(t, tl);
59                getnxt(t2, tl);
60                for(k=0; k<n; k++)  // 枚举所有文本串
61                    if(KMP(s[k], t, len[k], tl) == -1 && KMP(s[k], t2, len[k], tl) == -1)
62                        break;
63                if(k>=n)    // 满足条件
64                    ans = max(ans, tl);
65            }
66        }
67        printf("%d\n", ans);
68    }
69    return 0;
70}