POJ 3080 Blue Jeans——暴力 + kmp

题意：找n个串都包含的最长连续子串，若子串的长度小于三输出no ，若有多个子串输出字典序最小的

思路：暴力枚举第一个串的全部连续子串（O（n^2）），然后用这个子串和其他字符串进行kmp

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 100;
char str[maxn][maxn], temp[maxn], ans[maxn];
int T, n, Next[maxn];
void makeNext(const char *p, int len) {
    int i = 0, j = 0;
    Next[i] = j;
    for (i = 1; i < len; i++) {
        while (j && p[i] != p[j]) j = Next[j - 1];
        if (p[i] == p[j]) j++;
        Next[i] = j;
    }
}
bool kmp(const char *s, const char *p, int len) {
    makeNext(p, len);
    for (int i = 0, j = 0; i < 60; i++) {
        while (j && s[i] != p[j]) j = Next[j - 1];
        if (s[i] == p[j]) j++;
        if (j == len) {
            return true;
        }
    }
    return false;
}
int main() {
    scanf("%d", &T);
    while (T--) {
        memset(ans, 0, sizeof(ans));
        int len = 0;
        scanf("%d", &n);
        for (int i = 0; i < n; i++) {
            scanf("%s", str[i]);
        }
        for (int i = 0; i < 60; i++) {
            for (int j = i; j < 60; j++) {
                int cnt = 0;
                for (int k = i; k <= j; k++) {
                    temp[cnt++] = str[0][k];
                }
                temp[cnt] = '\0';
                bool ok = true;
                for (int k = 1; k < n; k++) {
                    if (!kmp(str[k], temp, j - i + 1)) {
                        ok = false; break;
                    }
                }
                if (ok) {
                    if ((len < j - i + 1) || (len == j - i + 1 && strcmp(ans, temp) > 0)) {
                        len = j - i + 1;
                        strcpy(ans, temp);
                    }
                }
            }
        }
        if (len < 3) {
            printf("no significant commonalities\n");
        }
        else {
            printf("%s\n", ans);
        }
    }
    return 0;
}