POJ-3080-Blue jeans(KMP, 暴力)

链接:

https://vjudge.net/problem/POJ-3080#author=alexandleo

题意:

给你一些字符串，让你找出最长的公共子串。

思路:

暴力枚举第一个串的子串,挨个匹配接下来的所有串.

找出最长子串中, 字典序最小的.

代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
//#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
#include <string>
#include <assert.h>
#include <iomanip>
#include <iostream>
#include <sstream>
#define MINF 0x3f3f3f3f
using namespace std;
typedef long long LL;
const int MAXN = 1e6+10;

int Next[MAXN];
string s[20];
int n;

void GetNext(string p)
{
    int len = p.length();
    Next[0] = -1;
    int j = 0;
    int k = -1;
    while (j < len-1)
    {
        if (k == -1 || p[k] == p[j])
        {
            ++k;
            ++j;
            Next[j] = k;
        }
        else
            k = Next[k];
    }
}

bool Kmp(string ss, string p)
{
    int lens = ss.length();
    int lenp = p.length();
    int i = 0, j = 0;
    while (i < lens && j < lenp)
    {
        if (j == -1 || ss[i] == p[j])
        {
            i++;
            j++;
        }
        else
            j = Next[j];
    }
    return j == lenp;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--)
    {
        string res = "";
        cin >> n;
        for (int i = 1;i <= n;i++)
            cin >> s[i];
        int len = s[1].length();
        for (int i = 0;i < len;i++)
        {
            for (int j = i;j < len;j++)
            {
                string tmp(s[1], i, j-i+1);
                GetNext(tmp);
                bool flag = true;
                for (int k = 2;k <= n;k++)
                {
                    if (!Kmp(s[k], tmp))
                    {
                        flag = false;
                        break;
                    }
                }
                if (flag)
                {
                    if (tmp.length() > res.length())
                        res = tmp;
                    if (tmp.length() == res.length() && tmp < res)
                        res = tmp;
                }
            }
        }
        if (res.length() >= 3)
            cout << res << endl;
        else
            cout << "no significant commonalities" << endl;
    }

    return 0;
}