python/TangDou/Topic/P3375.cpp

#include <bits/stdc++.h>

using namespace std;
const int N = 1000010;
int n, m, ne[N];
char s[N], p[N];

int main() {
#ifndef ONLINE_JUDGE
    freopen("P3375.in", "r", stdin);
#endif
    cin >> (s + 1) >> (p + 1);            // 先长串，再短串
    n = strlen(p + 1), m = strlen(s + 1); // 自已来测长

    // 求模式串ne数组
    for (int i = 2, j = 0; i <= n; i++) {
        while (j && p[i] != p[j + 1]) j = ne[j];
        if (p[i] == p[j + 1]) j++;
        ne[i] = j;
    }

    // 求源串中模式串出现的每个位置
    for (int i = 1, j = 0; i <= m; i++) {
        while (j && s[i] != p[j + 1]) j = ne[j];
        if (s[i] == p[j + 1]) j++;
        if (j == n) {
            printf("%d\n", i - n + 1);
            j = ne[j]; // 继续搜索，重置 j=ne[j]
        }
    }
    // 本题要求最后输出模式串的ne数组
    for (int i = 1; i <= n; i++) cout << ne[i] << " ";
    return 0;
}