POJ 3080 Blue Jeans (kmp)

题意：

求最长的公共子串

思路：

枚举第一个串的后缀串，采用kmp与后面的串进行匹配。至于为什么不用枚举完所有的子串，因为kmp的过程中已经有过子串的匹配过程

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <string>



using namespace std;



string str[20];

int p[64];



void getp(string b)

{

    p[0] = -1;

    int j = -1;

    for (int i = 1; i < b.size(); ++i)

    {

        while (j != -1 && b[j+1] != b[i])

            j = p[j];

        if (b[j+1] == b[i])

            ++j;

        p[i] = j;

    }

}



int kmp(string a, string b)         // 返回b能匹配到a的最大长度

{

    getp(b);

    int j = -1;

    int ans = -1;

    for (int i = 0; i < a.size(); ++i)

    {

        while (j != -1 && b[j+1] != a[i])

            j = p[j];

        if (b[j+1] == a[i])

            ++j;

        if (j > ans)

            ans = j;

        if (j+1 == b.size())

            break;

    }

    return ans;

}



int main()

{

    int cases;

    cin >> cases;

    while (cases--)

    {

        int n, ans = 0;

        cin >> n;

        

        for (int i = 0; i < n; ++i)

            cin >> str[i];

        

        string outprint;



        for (int i = 0; i < 60; ++i)

        {

            string a, b;

            b = str[0].substr(i);

            int minmax = 100;



            for (int j = 1; j < n; ++j)     // 求后缀串能匹配到其它字符串的最大长度

            {

                a = str[j];

                int temp = kmp(a, b) + 1;

                if (temp < minmax)

                    minmax = temp;

            }

            if (minmax > ans)

                ans = minmax, outprint = str[0].substr(i, minmax);

            else if (minmax == ans && str[0].substr(i, minmax) < outprint)

                outprint = str[0].substr(i, minmax);

        }

        if (ans < 3) 

            cout << "no significant commonalities" << endl;

        else 

            cout << outprint << endl;

    }

    return 0;

}