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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#517146#1281. Longest Common SubsequenceZ_drjTL 7ms39124kbC++141.9kb2024-08-13 09:20:512024-08-13 09:20:51

Judging History

你现在查看的是最新测评结果

  • [2024-08-13 09:20:51]
  • 评测
  • 测评结果:TL
  • 用时:7ms
  • 内存:39124kb
  • [2024-08-13 09:20:51]
  • 提交

answer

#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>

using i64 = long long;

const int N = 1e6 + 5;
const int INF = 2E6;

int n, m;
int a[N], b[N];
int sa[N], sb[N];

struct FenwickTree {
	int t[N << 2];
	void clear() {
		std::fill(t + 1, t + 2 * INF + 2, -INF);
	}
	int lowbit(int x) {
		return x & -x;
	}
	void update(int x, int val) {
		x += INF + 1;
		for (; x; x -= lowbit(x)) {
			t[x] = std::max(t[x], val);
		}
	}
	int query(int x) {
		x += INF + 1;
		int ans = 0;
		for (; x <= INF * 2 + 1; x += lowbit(x)) {
			ans = std::max(ans, t[x]);
		}
		return ans;
	}
}t1, t2;

void solve() {
	scanf("%d %d", &n, &m);

	for (int i = 1; i <= n; i++) {
		scanf("%d", a + i);
		sa[i] = sa[i - 1] + (a[i] == 2);
	}

	for (int i = 1; i <= m; i++) {
		scanf("%d", b + i);
		sb[i] = sb[i - 1] + (b[i] == 2);
	}

	sa[n + 1] = sa[n], sb[m + 1] = sb[m];

	std::vector<int> la, ra, lb, rb;
	la.push_back(0), lb.push_back(0);
	ra.push_back(n + 1), rb.push_back(m + 1);

	for (int i = 1; i <= n; i++) {
		if (a[i] == 1) {
			la.push_back(i);
		}
	}

	for (int i = n; i >= 1; i--) {
		if (a[i] == 3) {
			ra.push_back(i);
		}
	}

	for (int i = 1; i <= m; i++) {
		if (b[i] == 1) {
			lb.push_back(i);
		}
	}

	for (int i = m; i >= 1; i--) {
		if (b[i] == 3) {
			rb.push_back(i);
		}
	}

	t1.clear(), t2.clear();

	int ans = std::min(sa[n], sb[m]);
	for (int i = std::min((int)la.size(), (int)lb.size()) - 1, j = 0; i >= 0; i--) {
		while (j < std::min((int)ra.size(), (int)rb.size()) && la[i] < ra[j] && lb[i] < rb[j]) {
			t1.update(sa[ra[j]] - sb[rb[j]], j + sb[rb[j]]), t2.update(sb[rb[j]] - sa[ra[j]], j + sa[ra[j]]), ++j;
		}
		ans = std::max(ans, i + t1.query(sa[la[i]] - sb[lb[i]]) - sb[lb[i]]);
		ans = std::max(ans, i + t2.query(sb[lb[i]] - sa[la[i]]) - sa[la[i]]);
	}

	printf("%d\n", ans);
}

int main() {
	int T; scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 7ms
memory: 39124kb

input:

3
3 3
1 2 3
1 2 3
3 3
1 1 1
1 1 2
3 3
1 3 2
1 2 3

output:

3
2
2

result:

ok 3 tokens

Test #2:

score: -100
Time Limit Exceeded

input:

139889
1 10
2
2 1 2 2 3 1 1 2 3 3
7 1
3 2 3 3 1 1 2
2
6 1
2 1 3 1 1 1
1
8 7
3 1 3 3 2 2 3 1
3 2 2 1 3 3 3
10 3
3 2 1 1 2 2 1 1 1 1
3 1 1
5 2
1 2 1 3 1
1 2
1 4
1
3 3 3 3
7 2
3 1 2 1 2 2 3
3 2
6 2
3 1 1 2 1 3
1 3
1 4
1
3 1 1 3
4 2
2 3 2 3
1 3
1 8
1
1 1 3 1 3 3 3 1
4 1
1 3 3 3
1
10 10
3 1 2 1 2 2 2 2 1...

output:


result: