Submission #133087

#	Time	Username	Problem	Language	Result	Execution time	Memory
133087		E869120	Necklace (Subtask 1-3) (BOI19_necklace1)	C++14	85 / 85	485 ms	71416 KiB

necklace

#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <tuple>
using namespace std;

int N, M, pa[3009][3009], pb[3009][3009];
int EA[6009], EB[6009];

tuple<int, int, int> solve(string S, string T) {
	N = S.size(); M = T.size();
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < M; j++) {
			pa[i][j] = 0;
			if (S[i] == T[j]) pa[i][j] = 1;
			if (i >= 1 && j >= 1 && S[i] == T[j]) {
				pa[i][j] = pa[i - 1][j - 1] + 1;
			}
		}
	}
	for (int i = N - 1; i >= 0; i--) {
		for (int j = M - 1; j >= 0; j--) {
			pb[i][j] = 0;
			if (S[i] == T[j]) pb[i][j] = 1;
			if (i < N - 1 && j < M - 1 && S[i] == T[j]) {
				pb[i][j] = pb[i + 1][j + 1] + 1;
			}
		}
	}

	int maxn = 0; pair<int, int> maxid = make_pair(-1, -1);
	for (int i = 0; i <= N; i++) {
		// S における境界が i の場合
		vector<int> A, B;
		for (int j = 0; j < N; j++) {
			if (i == N) A.push_back(0);
			else A.push_back(pb[i][j]);
		}
		for (int j = 0; j < N; j++) {
			if (i == 0) B.push_back(0);
			else B.push_back(pa[i - 1][j]);
		}
		for (int j = 0; j <= N * 2; j++) { EA[j] = (1 << 29); EB[j] = -(1 << 29); }
		for (int j = 0; j < N; j++) {
			EA[A[j] + j] = min(EA[A[j] + j], j);
			EB[max(0, -B[j] + j + 1)] = max(EB[max(0, -B[j] + j + 1)], j);
		}
		
		// ret1 には条件を満たす最も左のインデックスを記録
		int ret1 = -(1 << 29);
		for (int j = 0; j <= N * 2; j++) {
			ret1 = max(ret1, EB[j]);
			int cr = ret1, cl = EA[j];
			if (maxn < cr - cl + 1) {
				maxn = cr - cl + 1;
				maxid = make_pair(i - B[cr], cl);
			}
		}
	}
	return make_tuple(maxn, maxid.first, maxid.second);
}

int main() {
	string A1, B1, B2; cin >> A1 >> B1; B2 = B1; reverse(B2.begin(), B2.end());
	tuple<int, int, int> v1 = solve(A1, B1);
	tuple<int, int, int> v2 = solve(A1, B2);

	cout << max(get<0>(v1), get<0>(v2)) << endl;
	if (get<0>(v1) > get<0>(v2)) cout << get<1>(v1) << " " << get<2>(v1) << endl;
	if (get<0>(v1) <= get<0>(v2)) cout << get<1>(v2) << " " << M - (get<2>(v2) + get<0>(v2)) << endl;
	return 0;
}

• Subtask 1: 25 / 25
• Subtask 2: 20 / 20
• Subtask 3: 40 / 40