[Cometoj#4 E]公共子序列_贪心_树状数组_动态规划

公共子序列

题目链接：https://cometoj.com/contest/39/problem/E?problem_id=1585

数据范围：略。

---

题解：

首先可以考虑知道了$1$的个数和$3$的个数，怎么求？

其实就是从开始找$x$个$1$，从结尾找$z$个$3$，然后两个序列中间$2$的个数的较小值。

然后按照官方题解那样推式子，发现可以用树状数组维护。

代码：

#include <bits/stdc++.h>

#define N 5000010 

using namespace std;

char *p1, *p2, buf[100000];

#define nc() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1 ++ )

int rd() {
	int x = 0;
	char c = nc();
	while (c < 48) {
		c = nc();
	}
	while (c > 47) {
		x = (((x << 2) + x) << 1) + (c ^ 48), c = nc();
	}
	return x;
}

struct BT {
	int tree[N << 1], n;

	inline int lowbit(int x) {
		return x & (-x);
	}

	void update(int x, int val) {
		for (int i = x; i <= n; i += lowbit(i)) {
			tree[i] = max(tree[i], val);
		}
	}

	int query(int x) {
		int ans = -999999999;
		for (int i = x; i; i -= lowbit(i)) {
			ans = max(ans, tree[i]);
		}
		return ans;
	}
}T1, T2;

struct Node {
	int x, y;
}st1[N], st2[N];

int a[N], b[N], sa[N], sb[N], cnt1, cnt2;

int main() {
	// cout << (int)0xefefefef << ' ' << -999999999 << endl ;
	int T = rd();
	while (T -- ) {
		int n = rd(), m = rd();
		T1.n = T2.n = n + m + 1;
		for (int i = 1; i <= n + m + 1; i ++ ) {
			T1.tree[i] = T2.tree[i] = -999999999;
		}
		for (int i = 1; i <= n; i ++ ) {
			a[i] = rd();
			sa[i] = sa[i - 1];
			if (a[i] == 2) {
				sa[i] ++ ;
			}
		}
		for (int i = 1; i <= m; i ++ ) {
			b[i] = rd();
			sb[i] = sb[i - 1];
			if (b[i] == 2) {
				sb[i] ++ ;
			}
		}

		int u1 = 1, u2 = 1;
		cnt1 = 0;
		while (1) {
			for (; a[u1] != 1 && u1 < n; u1 ++ );
			for (; b[u2] != 1 && u2 < m; u2 ++ );
			if (a[u1] != 1 || b[u2] != 1) {
				break;
			}
			st1[ ++ cnt1] = (Node) {u1, u2};
			u1 ++ , u2 ++ ;
		}
		u1 = n, u2 = m;
		cnt2 = 0;
		while (1) {
			for (; a[u1] != 3 && u1 > 1; u1 -- );
			for (; b[u2] != 3 && u2 > 1; u2 -- );
			if (a[u1] != 3 || b[u2] != 3) {
				break;
			}
			st2[ ++ cnt2] = (Node) {u1, u2};
			u1 -- , u2 -- ;
		}
		int now = 0, ans = 0;
		st2[0] = (Node) {n, m};
		for (int i = cnt1; ~i; i -- ) {
			int x = st1[i].x, y = st1[i].y, mdl;
			for (; st2[now].x >= x && st2[now].y >= y && now <= cnt2; now ++ ) {
				mdl = sa[st2[now].x] - sb[st2[now].y];
				T1.update(m + 1 + mdl, now + sa[st2[now].x]);
				T2.update(n + 1 - mdl, now + sb[st2[now].y]);
			}
			mdl = sa[x] - sb[y];
			ans = max(ans, i + T1.query(m + 1 + mdl) - sa[x]);
			ans = max(ans, i + T2.query(n + 1 - mdl) - sb[y]);
		}
		printf("%d\n", ans);
	}
	return 0;
}