【kmp算法next数组应用】Simpsons’ Hidden Talents HDU

Think:
1知识点：kmp算法next数组应用
2题意：输入两个串，询问“最长前后缀”，“最长前后缀”定义为：串ans为st1的前缀，st2的后缀，输出“最长前后缀”和“最长前后缀”的长度
3解题思路：连接串st1和st2，求next数组，next[len-1]表示串st的最大公共前后缀，需要判断len与len_st1和len_st2的长度，若len > len_st1 或 len > len_st2，则len = next[len-1];
4反思：
（1）：next数组开小了
（2）：strcar(st1, st2)函数拼接串st1和串st2时要求st1数组的内存存储空间可以存储长度为len_st1+len_st2的串;

以下为Time Limit Exceeded代码——next数组开小了以及st1数组开小了

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int max_size = 50414;int _next[max_size];char st1[max_size], st2[max_size];void get_next(char *P, int len_P);int main(){    while(~scanf("%s %s", st1, st2)){        int len1 = strlen(st1);        int len2 = strlen(st2);        strcat(st1, st2);        int len = len1 + len2;        get_next(st1, len);        int cir_len = _next[len-1];        while(cir_len > len1 || cir_len > len2){            cir_len = _next[cir_len-1];        }        for(int i = 0; i < cir_len ; i++)            printf("%c", st1[i]);        if(cir_len) printf(" ");        printf("%d\n", cir_len);    }    return 0;}void get_next(char *P, int len_P){    int q, k;    _next[0] = 0;    k = 0;    for(q = 1; q < len_P; q++){        while(k > 0 && P[q] != P[k]){            k = _next[k-1];        }        if(P[q] == P[k])            k++;        _next[q] = k;    }    return;}

以下为Accepted代码

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int size_T = 50414;const int size_P = size_T << 1;int _next[size_P];char st1[size_P], st2[size_T];void get_next(char *P, int len_P);int main(){    while(~scanf("%s %s", st1, st2)){        int len1 = strlen(st1);        int len2 = strlen(st2);        strcat(st1, st2);        int len = len1 + len2;        get_next(st1, len);        int cir_len = _next[len-1];        while(cir_len > len1 || cir_len > len2){            cir_len = _next[cir_len-1];        }        for(int i = 0; i < cir_len ; i++)            printf("%c", st1[i]);        if(cir_len) printf(" ");        printf("%d\n", cir_len);    }    return 0;}void get_next(char *P, int len_P){    int q, k;    _next[0] = 0;    k = 0;    for(q = 1; q < len_P; q++){        while(k > 0 && P[q] != P[k]){            k = _next[k-1];        }        if(P[q] == P[k])            k++;        _next[q] = k;    }    return;}