AtCoder Beginner Contest 207 F Tree Patrolling

洛谷传送门

AtCoder 传送门

简单树形 dp。

设 \(f_{u,i,p=0/1,q=0/1}\) 为 \(u\) 的子树中被覆盖点数为 \(i\)，\(u\) 有没有被覆盖，\(u\) 有没有被选。

转移树形背包合并即可，需要分类讨论。要注意如果 \(u\) 没被覆盖，\(v\) 选了，或者 \(u\) 选了，\(v\) 没被覆盖，被覆盖点数要 \(+1\)。

式子较复杂，具体见代码。

code

// Problem: F - Tree Patrolling
// Contest: AtCoder - AtCoder Beginner Contest 207
// URL: https://atcoder.jp/contests/abc207/tasks/abc207_f
// Memory Limit: 1024 MB
// Time Limit: 3000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair<ll, ll> pii;

const int maxn = 2020;
const ll mod = 1000000007;

ll n, f[maxn][maxn][2][2], sz[maxn], head[maxn], len, g[maxn][2], h[maxn][maxn][2];
struct edge {
	int to, next;
} edges[maxn << 1];

inline void add_edge(int u, int v) {
	edges[++len].to = v;
	edges[len].next = head[u];
	head[u] = len;
}

void dfs(int u, int fa) {
	f[u][0][0][0] = f[u][1][1][1] = 1;
	sz[u] = 1;
	for (int i = head[u]; i; i = edges[i].next) {
		int v = edges[i].to;
		if (v == fa) {
			continue;
		}
		dfs(v, u);
		for (int j = 0; j <= sz[u]; ++j) {
			g[j][0] = f[u][j][0][0];
			g[j][1] = f[u][j][1][0];
			f[u][j][0][0] = f[u][j][1][0] = 0;
		}
		for (int j = 0; j <= sz[u]; ++j) {
			for (int k = 0; k <= sz[v]; ++k) {
				for (int x = 0; x <= 1; ++x) {
					for (int y = 0; y <= 1; ++y) {
						if (!x && y) {
							f[u][j + k][0][0] = (f[u][j + k][0][0] + g[j][0] * f[v][k][1][0] % mod) % mod;
							f[u][j + k + 1][1][0] = (f[u][j + k + 1][1][0] + g[j][0] * f[v][k][1][1] % mod) % mod;
						} else {
							f[u][j + k][x | y][0] = (f[u][j + k][x | y][0] + g[j][x] * h[v][k][y] % mod) % mod;
						}
					}
				}
			}
		}
		for (int j = 0; j <= sz[u]; ++j) {
			g[j][0] = f[u][j][0][1];
			g[j][1] = f[u][j][1][1];
			f[u][j][0][1] = f[u][j][1][1] = 0;
		}
		for (int j = 0; j <= sz[u]; ++j) {
			for (int k = 0; k <= sz[v]; ++k) {
				for (int x = 0; x <= 1; ++x) {
					f[u][j + k + (x ^ 1)][1][1] = (f[u][j + k + (x ^ 1)][1][1] + g[j][1] * h[v][k][x] % mod) % mod;
				}
			}
		}
		sz[u] += sz[v];
	}
	for (int i = 0; i <= sz[u]; ++i) {
		h[u][i][0] = (f[u][i][0][0] + f[u][i][0][1]) % mod;
		h[u][i][1] = (f[u][i][1][0] + f[u][i][1][1]) % mod;
	}
}

void solve() {
	scanf("%lld", &n);
	for (int i = 1, u, v; i < n; ++i) {
		scanf("%d%d", &u, &v);
		add_edge(u, v);
		add_edge(v, u);
	}
	dfs(1, -1);
	for (int i = 0; i <= n; ++i) {
		printf("%lld\n", (h[1][i][0] + h[1][i][1]) % mod);
	}
}

int main() {
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}