昨晚与集训队的诸位聚餐，得悉弘毅的选拔比预想中要近，而且英语入学考也会与是否大一能参加四级考有关。结束后，第一次来到武大ACM训练室，被一桌论文、草稿、书籍、KFC、外卖袋、电脑、可乐瓶……的混沌空间深深震撼。黑暗的夜幕下，机房灯火闪烁，两个大三巨佬与一个大一萌新，就这样整整齐齐地坐成一排，玩起了音游……返程时，我白嫖了亿本本资料书，md晚上聚餐花了200多还没饱也算时不枉此行吧:)
- \(kmp\)
神秘の\(DFA\)
luoguP3375【模板】KMP字符串匹配
# include "bits/stdc++.h"
using namespace std;
constexpr int N = 1e6 + 3;
char txt[N + 1], pat[N + 1];
int main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
// cin >> (txt + 1) >> (pat + 1); // in luogu, this sentence will lead to RE
scanf("%s%s", txt + 1, pat + 1); // for convenience and to prevent the array from going out of bounds, the srings subscript starts at 1
int len_1 = strlen(txt + 1), len_2 = strlen(pat + 1);
vector<int> nxt(len_1 + 1);
int j = 0;
for(int i = 2; i <= len_2; ++i) { // the pattern string will be matched by itself to obtain the nxt array that conveniently makes the prefix and suffix the same
while(j && pat[i] != pat[j + 1]) j = nxt[j]; // seek forward for j with the same prefix
if(pat[i] == pat[j + 1]) ++j;
nxt[i] = j; // hwo should i jump after i + 1 mismatch
}
j = 0; // j can be viewed as the position of the last place of the pattern string that has currently been matched
for(int i = 1; i <= len_1; ++i) {
while(j && txt[i] != pat[j + 1]) j = nxt[j]; // if it fails in matching, it keeps jumping back until it can match again
if(txt[i] == pat[j + 1]) ++j; // if the match is successful, the corresponding pattern string position is advanced one place
if(j == len_2) {
printf("%d\n", i - j + 1);
j = nxt[j];
}
}
for(int i = 1; i <= len_2; ++i) printf("%d ", nxt[i]);
return 0;
}
