洛谷 P8085 [COCI2011-2012#4] KRIPTOGRAM

洛谷传送门

思路

注意到只要字符串出现的相对位置匹配就行。设 \(a_i,b_i\) 分别为明文/密文中第 \(i\) 个字符串与上一个跟它相同的字符串隔了几个单词（若该字符串第一次出现则设为 \(+\infty\)）。然后直接 KMP。注意可能有这种情况：明文为 c (a b c)、密文为 x y z，此时 \(a_4 = 3\)，但 \(b_3 = +\infty\)。因此还要特判 \(b_i = +\infty\) 的情况，即如果 \(b_j = +\infty\) 且 \(a_i \ge j\) 时就能匹配。

代码

code

/*

p_b_p_b txdy
AThousandMoon txdy
AThousandSuns txdy
hxy txdy

*/

#include <bits/stdc++.h>
#define pb push_back
#define fst first
#define scd second

using namespace std;
typedef long long ll;
typedef pair<ll, ll> pii;

const int maxn = 1000100;
const int inf = 0x3f3f3f3f;

map<string, int> mp;
int n, m, a[maxn], b[maxn], fail[maxn];

void solve() {
	string s;
	while (cin >> s) {
		if (s[0] == '$') {
			break;
		}
		++n;
		if (mp.find(s) == mp.end()) {
			a[n] = inf;
			mp[s] = n;
		} else {
			a[n] = n - mp[s];
			mp[s] = n;
		}
	}
	mp.clear();
	while (cin >> s) {
		if (s[0] == '$') {
			break;
		}
		++m;
		if (mp.find(s) == mp.end()) {
			b[m] = inf;
			mp[s] = m;
		} else {
			b[m] = m - mp[s];
			mp[s] = m;
		}
	}
	for (int i = 2, j = 0; i <= m; ++i) {
		while (j && b[i] != b[j + 1]) {
			j = fail[j];
		}
		if (b[i] == b[j + 1]) {
			++j;
		}
		fail[i] = j;
	}
	for (int i = 1, j = 0; i <= n; ++i) {
		while (j && ((b[j + 1] == inf && a[i] < j + 1) || (b[j + 1] != inf && b[j + 1] != a[i]))) {
			j = fail[j];
		}
		if (!((b[j + 1] == inf && a[i] < j + 1) || (b[j + 1] != inf && b[j + 1] != a[i]))) {
			++j;
		}
		// printf("%d %d\n", i, j);
		if (j == m) {
			printf("%d\n", i - m + 1);
			return;
		}
	}
}

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}