HDU2594 Simpsons ' Hidden talents "KMP"

Topic Links:

http://acm.hdu.edu.cn/showproblem.php?pid=2594

Main topic:

Give you two strings of S1 and S2, which is the longest string and length of both the S1 prefix and the S2 suffix.

Ideas:

KMP algorithm next[j] = = k is the essence of the current letter mismatch, the pattern string of the first K (s0~sk-1) and position J before

K-item (SJ-1-K~SJ-1) is equal, this K-value is all that satisfies the above situation maximum. So the meaning of Next[len]

is the longest prefix of the pattern string and the length of the string equal to the suffix.

Using the nature of next[], the string S2 is connected to the S1 behind. Next[] Array for S1. Well, now the Next[len]

is the longest length of the S1 prefix and S2 suffix, which is only full when the length of the <= string S1 and the length of the S2

of the foot. If the above conditions are not met, the found string is not a substring of S1 or S2. Let Len = Next[len] continue to check

Find the longest length that satisfies the condition. Find the string that satisfies the maximum length of the request, output S0~s (len-1).

AC Code:

#include <iostream> #include <algorithm> #include <cstdio> #include <cstring>using namespace Std;char s1[100010],s2[50010];int next[100010],len,len1,len2;void GetNext (char *s) {    int i = 0,j =-1;    Next[0] =-1;    while (i <= len)    {        if (j = =-1 | | s[i] = = S[j])        {            i++,j++;            Next[i] = j;        }        else            j = next[j];    }} int main () {    while (cin >> s1 >> S2)    {        len1 = strlen (S1);        Len2 = strlen (s2);        strcat (S1,S2);        Len = len1 + len2;        GetNext (S1);        while (Next[len] > Len1 | | Next[len] > Len2)            len = Next[len];        len = Next[len];        for (int i = 0; i < len; ++i)            cout << s1[i];        if (len)            cout << ';        cout << len << endl;    }    return 0;}

HDU2594 Simpsons ' Hidden talents "KMP"