34 lines
956 B
C++
34 lines
956 B
C++
#include <bits/stdc++.h>
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using namespace std;
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const int N = 1000010;
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int n, m, ne[N];
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char s[N], p[N];
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int main() {
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#ifndef ONLINE_JUDGE
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freopen("P3375.in", "r", stdin);
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#endif
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cin >> (s + 1) >> (p + 1); // 先长串，再短串
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n = strlen(p + 1), m = strlen(s + 1); // 自已来测长
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// 求模式串ne数组
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for (int i = 2, j = 0; i <= n; i++) {
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while (j && p[i] != p[j + 1]) j = ne[j];
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if (p[i] == p[j + 1]) j++;
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ne[i] = j;
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}
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// 求源串中模式串出现的每个位置
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for (int i = 1, j = 0; i <= m; i++) {
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while (j && s[i] != p[j + 1]) j = ne[j];
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if (s[i] == p[j + 1]) j++;
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if (j == n) {
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printf("%d\n", i - n + 1);
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j = ne[j]; // 继续搜索，重置 j=ne[j]
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}
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}
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// 本题要求最后输出模式串的ne数组
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for (int i = 1; i <= n; i++) cout << ne[i] << " ";
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return 0;
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} |