[CF1065F]Up and Down the Tree_tarjan_树形dp

Up and Down the Tree

题目链接：https://www.luogu.org/problem/CF1065F

数据范围：略。

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题解：

我们把每个叶子向它上面$k$个点连边，然后trajan缩点。

表示如果一个$SCC$中的叶子能走到，剩下的就都能。

然后我们就求一个最长的根缀链即可。

代码：

#include <bits/stdc++.h>

#define setIO(s) freopen(s".in", "r", stdin), freopen(s".out", "w", stdout) 

#define N 1000010 

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, f = 1;
	char c = nc();
	while (c < 48) {
		if (c == '-')
			f = -1;
		c = nc();
	}
	while (c > 47) {
		x = (((x << 2) + x) << 1) + (c ^ 48), c = nc();
	}
	return x * f;
}

struct Graph {
	int head[N], to[N << 1], nxt[N << 1], tot;

	inline void add(int x, int y) {
		to[ ++ tot] = y;
		nxt[tot] = head[x];
		head[x] = tot;
	}
}G1, G2;

int f[21][N];

int dep[N], low[N], cnt, sz[N], F[N], st[N], top, blg[N];

bool vis[N], ins[N];

queue <int> q;

void tarjan(int p) {
	dep[p] = low[p] = ++cnt;
	vis[p] = ins[p] = true;
	st[ ++ top] = p;
	for (int i = G1.head[p]; i; i = G1.nxt[i]) {
		if (!vis[G1.to[i]]) {
			tarjan(G1.to[i]), low[p] = min(low[p], low[G1.to[i]]);
		}
		else if (ins[G1.to[i]]) {
			low[p] = min(low[p], dep[G1.to[i]]);
		}
	}
	if (dep[p] == low[p]) {
		blg[0] ++ ;
		int t;
		do {
			t = st[top -- ];
			blg[t] = blg[0];
			ins[t] = false;
		} while(t != p);
	}
}

bool lf[N];

void dfs(int p, int fa) {
	f[0][p] = fa;
	for (int i = 1; i <= 20; i ++ ) {
		f[i][p] = f[i - 1][f[i - 1][p]];
	}
	for (int i = G1.head[p]; i; i = G1.nxt[i]) {
		dfs(G1.to[i], p);
	}
}

int d[N];

int main() {
	// setIO("c");
	memset(lf, true, sizeof lf);
	int n = rd(), k = rd();
	for (int i = 2; i <= n; i ++ ) {
		int x = rd();
		G1.add(x, i);
		lf[x] = false;
	}
	dfs(1, 1);
	for (int i = 2; i <= n; i ++ ) {
		if (lf[i]) {
			int x = i;
			for (int j = 20; ~j; j -- ) {
				if (k & (1 << j)) {
					x = f[j][x];
				}
			}
			G1.add(i, x);
		}
	}
	tarjan(1);
	for (int i = 1; i <= n; i ++ ) {
		for (int j = G1.head[i]; j; j = G1.nxt[j]) {
			if (blg[i] != blg[G1.to[j]]) {
				G2.add(blg[i], blg[G1.to[j]]);
				d[blg[G1.to[j]]] ++ ;
			}
		}
	}

	for (int i = 2; i <= n; i ++ ) {
		if (lf[i]) {
			sz[blg[i]] ++ ;
		}
	}

	for (int i = 1; i <= blg[0]; i ++ ) {
		F[i] = sz[i];
	}

	q.push(blg[1]);
	while (!q.empty()) {
		int x = q.front();
		q.pop();
		for (int i = G2.head[x]; i; i = G2.nxt[i]) {
			F[G2.to[i]] = max(F[x] + sz[G2.to[i]], F[G2.to[i]]);
			d[G2.to[i]] -- ;
			if (!d[G2.to[i]]) {
				q.push(G2.to[i]);
			}
		}
	}

	int ans = 0;
	for (int i = 1; i <= blg[0]; i ++ ) {
		ans = max(ans, F[i]);
	}
	cout << ans << endl ;
	// fclose(stdin), fclose(stdout);
	return 0;
}