洛谷P4516 [JSOI2018] 潜入行动

朴素的设计 \(DP\)，\(dp_{i,j,0/1,0/1}\) 表示 \(i\) 的子树选了 \(j\) 个点，当前选没有，当前被覆盖没有。

令人火大的转移方程（具体转移边界还请看代码）：

\[\begin{align} &dp_{u,j,0,0} = \sum_{k=1}^{siz_u} dp_{u,k,0,0} \times dp_{v,j-k,0,1} \\ &dp_{u,j,0,1} = \begin{cases} \sum_{k=1}^{siz_u} dp_{u,k,0,0} \times dp_{v,j-k,1,1} \\ \sum_{k=1}^{siz_u} dp_{u,k,0,1} \times dp_{v,j-k,0,1} \\ \sum_{k=1}^{siz_u} dp_{u,k,0,1} \times dp_{v,j-k,1,1} \end{cases} \\ &dp_{u,j,1,0} = \begin{cases} \sum_{k=1}^{siz_u} dp_{u,k,1,0} \times dp_{v,j-k,0,0} \\ \sum_{k=1}^{siz_u} dp_{u,k,1,0} \times dp_{v,j-k,0,1} \end{cases} \\ &dp_{u,j,1,1} = \begin{cases} \sum_{k=1}^{siz_u} dp_{u,k,1,0} \times dp_{v,j-k,1,0} \\ \sum_{k=1}^{siz_u} dp_{u,k,1,0} \times dp_{v,j-k,1,1} \\ \sum_{k=1}^{siz_u} dp_{u,k,1,1} \times dp_{v,j-k,0,0} \\ \sum_{k=1}^{siz_u} dp_{u,k,1,1} \times dp_{v,j-k,0,1} \\ \sum_{k=1}^{siz_u} dp_{u,k,1,1} \times dp_{v,j-k,1,0} \\ \sum_{k=1}^{siz_u} dp_{u,k,1,1} \times dp_{v,j-k,1,1} \end{cases} \end{align} \]

最后注意转移边界以及外层倒序枚举就行了。

\(\Large \mathscr{Code}\)

#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 100, MOD = 1e9 + 7;
int n, m, dp[N][105][2][2], k, siz[N];
vector<int> g[N];
void dfs(int u, int fa) {
	siz[u] = dp[u][0][0][0] = dp[u][1][1][0] = 1;
	for (int e : g[u]) {
		if (e == fa) continue;
		dfs(e, u);
		siz[u] += siz[e];
		for (int i = min(k, siz[u]); i >= 0; i--) {
			int sum1 = 0, sum2 = 0, sum3 = 0, sum4 = 0;
			for (int j = 0; j <= min(i, siz[e]); j++) {
				if (i - j > siz[u] - siz[e]) continue;
				sum1 = (sum1 + 1ll * dp[u][i - j][0][0] * dp[e][j][0][1] % MOD) % MOD;

				sum2 = (sum2 + 1ll * dp[u][i - j][0][0] * dp[e][j][1][1] % MOD) % MOD;
				sum2 = (sum2 + 1ll * dp[u][i - j][0][1] * dp[e][j][0][1] % MOD) % MOD;
				sum2 = (sum2 + 1ll * dp[u][i - j][0][1] * dp[e][j][1][1] % MOD) % MOD;

				sum3 = (sum3 + 1ll * dp[u][i - j][1][0] * dp[e][j][0][0] % MOD) % MOD;
				sum3 = (sum3 + 1ll * dp[u][i - j][1][0] * dp[e][j][0][1] % MOD) % MOD;

				sum4 = (sum4 + 1ll * dp[u][i - j][1][0] * dp[e][j][1][0] % MOD) % MOD;
				sum4 = (sum4 + 1ll * dp[u][i - j][1][0] * dp[e][j][1][1] % MOD) % MOD;
				sum4 = (sum4 + 1ll * dp[u][i - j][1][1] * dp[e][j][0][0] % MOD) % MOD;
				sum4 = (sum4 + 1ll * dp[u][i - j][1][1] * dp[e][j][0][1] % MOD) % MOD;
				sum4 = (sum4 + 1ll * dp[u][i - j][1][1] * dp[e][j][1][0] % MOD) % MOD;
				sum4 = (sum4 + 1ll * dp[u][i - j][1][1] * dp[e][j][1][1] % MOD) % MOD;
			}
			dp[u][i][0][0] = sum1, dp[u][i][0][1] = sum2, dp[u][i][1][0] = sum3, dp[u][i][1][1] = sum4;
		}
	}
}
signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	cin >> n >> k;
	for (int i = 1; i < n; i++) {
		int x, y;
		cin >> x >> y;
		g[x].push_back(y), g[y].push_back(x);
	}
	dfs(1, 0);
	cout << (dp[1][k][0][1] + dp[1][k][1][1]) % MOD;
	return 0;
}