题解 P4815 [CCO2014] 狼人游戏

看题目限制，可以发现如果将机器人作为点，指控和保护关系作为边，可以建出一个森林，就下来就是传统的树形背包了。

设 $f_{i,j,0/1}$ 表示当前点为 $i$，子树内有 $j$ 个狼人，当前点是否为狼人的方案数。

初始化：$f_{u,0,0} = f_{u,1,1} = 1$

当前点为狼：

• 指控：$f_{u,j,1}=f_{u,j-k,1} \times f_{v,k,0}$
• 保护：$f_{u,j,1}=f_{u,j-k,1} \times f_{v,k,1}$

当前点为民：$f_{u,j,0}=f_{u,j-k,0} \times (f_{v,k,0}+f_{v,k,1})$

最后还需要把没棵树合并起来：$g_j=g_{j-k} \times (f_{i,k,0}+f_{i,k,1})$

code:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 200 + 5, mod = 1e9 + 7;
int n, num, m, vis[N];
ll f[N][N][2], tmp[N][2], g[N];
struct node{
	int v, w;
	node(int v = 0, int w = 0):v(v), w(w){}
};
vector<node> adj[N];
void dfs(int u) {
	vis[u] = 1;
	f[u][0][0] = f[u][1][1] = 1;
	for (int i = 0; i < adj[u].size(); ++i) dfs(adj[u][i].v);
	for (int i = 0; i < adj[u].size(); ++i) {
		int v = adj[u][i].v, w = adj[u][i].w;
		memset(tmp, 0, sizeof(tmp));
		for (int j = num; j >= 1; --j) {
			for (int k = 0; k < j; ++k) {
				tmp[j][1] = (tmp[j][1] + f[u][j - k][1] * f[v][k][w]) % mod;
			}
		}
		for (int j = num; j >= 1; --j) f[u][j][1] = tmp[j][1];
	}
	for (int i = 0; i < adj[u].size(); ++i) {
		int v = adj[u][i].v, w = adj[u][i].w;
		memset(tmp, 0, sizeof(tmp));
		for (int j = num; j >= 1; --j) {
			for (int k = 0; k <= j; ++k) {
				tmp[j][0] = (tmp[j][0] + f[u][j - k][0] * (f[v][k][0] + f[v][k][1])) % mod;
			}
		}
		for (int j = num; j >= 1; --j) f[u][j][0] = tmp[j][0];
	}
}
int main() {
	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	cin >> n >> num >> m;
	for (int i = 1; i <= m; ++i) {
		char type; int u, v;
		cin >> type >> u >> v;
		adj[u].push_back(node(v, (type == 'D')));
	}
	g[0] = 1;
	for (int i = 1; i <= n; ++i) {
		if (!vis[i]) {
			dfs(i);
			memset(tmp, 0, sizeof(tmp));
			for (int j = num; j >= 1; --j) {
				for (int k = 0; k <= j; ++k) {
					tmp[j][0] = (tmp[j][0] + g[j - k] * (f[i][k][0] + f[i][k][1])) % mod;
				}
			}
			for (int j = num; j >= 1; --j) g[j] = tmp[j][0];
		}
	}
	cout << g[num] << endl;
	return 0;
}