【kmp算法】Substrings HDU

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Think:
1知识点：kmp算法
2题意：T组测试数据，每组测试数据输入n个字符串，询问最长长度的子串x，要求子串x或者子串x的翻转串在每个字符串中都会出现
3思路：找到最小的字符串，然后暴力枚举子串查询，剪枝优化
4反思：
（1）：字符串长度为1的字符串不要忘记

vjudge题目链接

以下为Wrong Answer代码——未枚举字符串长度为1的字符串

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int inf = 0x3f3f3f3f;const int max_size = 144;int _next[max_size], len[max_size];char st[104][max_size], st1[max_size], st2[max_size];void get_next(char *P, int len_P);bool kmp(char *T, int len_T, char *P, int len_P);int main(){    int T, n, i, j, k, tmp, t;    scanf("%d", &T);    while(T--){        scanf("%d", &n);        tmp = inf;        for(i = 0; i < n; i++){            scanf("%s", st[i]);            len[i] = strlen(st[i]);            if(len[i] < tmp){                tmp = len[i], t = i;            }        }        strcpy(st1, st[t]);        int ans = 0;        for(i = 0; i < tmp; i++){            for(j = i+1; j < tmp; j++){                if(j-i+1 <= ans) continue;                t = 0;                for(k = j; k >= i; k--){                    st2[t++] = st1[k];                }                for(k = 0; k < n; k++){                    if(!kmp(st[k], len[k], &st1[i], j-i+1) && !kmp(st[k], len[k], st2, j-i+1))                        break;                }                if(k == n) ans = j-i+1;            }        }        printf("%d\n", ans);    }    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;}bool kmp(char *T, int len_T, char *P, int len_P){    int k, i;    get_next(P, len_P);    k = 0;    for(i = 0; i < len_T; i++){        while(k > 0 && P[k] != T[i]){            k = _next[k-1];        }        if(P[k] == T[i])            k++;        if(k == len_P) return true;    }    return false;}

以下为Accepted代码

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int inf = 0x3f3f3f3f;const int max_size = 144;int _next[max_size], len[max_size];char st[104][max_size], st1[max_size], st2[max_size];void get_next(char *P, int len_P);bool kmp(char *T, int len_T, char *P, int len_P);int main(){    int T, n, i, j, k, tmp, t;    scanf("%d", &T);    while(T--){        scanf("%d", &n);        tmp = inf;        for(i = 0; i < n; i++){            scanf("%s", st[i]);            len[i] = strlen(st[i]);            if(len[i] < tmp){                tmp = len[i], t = i;            }        }        strcpy(st1, st[t]);        int ans = 0;        for(i = 0; i < tmp; i++){            for(j = i; j < tmp; j++){                if(j-i+1 <= ans) continue;                t = 0;                for(k = j; k >= i; k--){                    st2[t++] = st1[k];                }                for(k = 0; k < n; k++){                    if(!kmp(st[k], len[k], &st1[i], j-i+1) && !kmp(st[k], len[k], st2, j-i+1))                        break;                }                if(k == n) ans = j-i+1;            }        }        printf("%d\n", ans);    }    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;}bool kmp(char *T, int len_T, char *P, int len_P){    int k, i;    get_next(P, len_P);    k = 0;    for(i = 0; i < len_T; i++){        while(k > 0 && P[k] != T[i]){            k = _next[k-1];        }        if(P[k] == T[i])            k++;        if(k == len_P) return true;    }    return false;}

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