HDU-2594 Simpsons’ Hidden Talents

解题思路:

法一:首先想到的是直接连接两个串，然后计算Next[]，如果Next[Len] > 任意一个串的长度的化，回溯直到小于等于两个串的长度

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <cstdio>
#include <vector>
#define MAXSIZE 1000100
using namespace std;

string a;
int Next[MAXSIZE];
int Len;

int GetNext()
{
    int i = 0, j = Next[0] = -1;
    while (i < Len)
    {
        if (j == -1 || a[i] == a[j])
            Next[++i] = ++j;
        else
            j = Next[j];
    }
}

int main(void)
{
    ios::sync_with_stdio(false);
    string t1, t2;
    while (cin >> t1)
    {
        cin >> t2;
        a = t1 + t2;
        Len = a.size();
        GetNext();
        if (Next[Len] == 0)
            cout << 0 << endl;
        else if (Next[Len] <= t1.size() && Next[Len] <= t2.size())    
            cout << a.substr(0, Next[Len]) << ' '<< Next[Len] << endl;
        else
        {
            int p = Next[Len];
            //backtracking, until find smaller than t1 and t2
            while (p > t1.size() || p > t2.size())
                p = Next[p];
            cout << a.substr(0, p) << ' ' << p << endl;
        }
    }

    return 0;
}

法二:把第二个串当做母串，第一个串当做模式串，注:要比较到第二个串的尾部，当模式串(第一个串)全部比对成功后，如果没到尾部，扔要比对，直到尾部

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#define N 50010
using namespace std;

char a[N];
char b[N];
int Next[N];
int Len_a, Len_b;

void GetNext()
{
    int i = 0, j = Next[0] = -1;
    while (i < Len_a)
    {
        if (j == -1 || a[i] == a[j])
            Next[++i] = ++j;
        else
            j = Next[j];
    }
}

int KMP()
{
    int i = 0, j = 0;
    GetNext();
    while (i < Len_b)
    {
        if (j == -1 || b[i] == a[j])
        {
            i++;
            j++;
        }    
        else
            j = Next[j];    
    }
    
    return j;
}

int main(void)
{
    ios::sync_with_stdio(false);
    while (cin >> a >> b)
    {
        Len_a = strlen(a);
        Len_b = strlen(b);
        int res = KMP();
        if (res)
        {
            for (int i = 0; i < res; ++i)
                cout << a[i];
            cout << ' ';
        }
        cout << res << endl;
    }
    
    return 0;
}