Description
自己看吧= =
Solution
这个题感觉很蛋疼啊= =因为它不仅仅是一棵树，在1节点处还有一个环。我们考虑一个环上距离节点1距离为dep对答案所做的贡献
假设环的长度为l,则贡献为
c1?kdep?(1+kl+k2l+....)
，然后我们惊讶的发现这原来是个等比数列求和啊。。。
于是对答案的贡献是
c1?kdep1?kl
,显然，修改m次的话我们最好把每个节点都指向1，这样dep=1，收益最大
之前一直没想明白环怎么办，看别人题解发现直接一个等比数列求和就搞定了= =too young。。
然后用
f[i][j][k]
表示节点i为根的子树，深度为j，修改k次获得最大值，先枚举1处的环长度再树dp即可，详情看代码
Code
#include
using namespace std; typedef long long LL; #define pb push_back #define mp make_pair #define F first #define S second inline void read(int &t) { int f = 1;char c; while (c = getchar(), c < '0' || c >
'9') if (c == '-') f = -1; t = c - '0'; while (c = getchar(), c >= '0' && c <= '9') t = t * 10 + c - '0'; t *= f; } const int N = 65; const double inf = 1e10; vector
g[N]; int tot, n, m, fa[N], cir[N], num[N]; double k, p[N], c[N], f[N][N][N]; void gao(int u, int dep, int len) { for (int i = 0, v; i < g[u].size(); ++i) { v = g[u][i]; gao(v, dep + 1, len); } for (int i = 0; i <= dep; ++i) { if (num[u] == len && i != 1) continue; if (u != 1 && num[u] && num[u] <= len && i != (len - (tot - dep))) continue; f[u][i][0] = c[u] * p[i] / (1.0 - p[len]); for (int j = 0, v; j < g[u].size(); ++j) { v = g[u][j]; for (int k = m; k >= 0; --k) { if (!k) f[u][i][k] += f[v][i + 1][k]; else { double t = -inf; for (int l = 0; l <= k; ++l) f[u][i][k] = max(f[u][i][k], f[u][i][k - l] + max(f[v][i + 1][l], l ? f[v][1][l - 1] : -inf)); f[u][i][k] = t; } } } } } int main() { read(n), read(m); scanf("%lf", &k); p[0] = 1.0; for (int i = 1; i <= n; ++i) p[i] = p[i - 1] * k; for (int i = 1; i <= n; ++i) { read(fa[i]); if (i != 1) g[fa[i]].pb(i); } for (int i = 1; i <= n; ++i) scanf("%lf", &c[i]); cir[++tot] = 1, num[1] = 1; int now = fa[1]; while (now != 1) { cir[++tot] = now; num[now] = tot; now = fa[now]; } double ans = 0.0;www.2cto.com for (int i = tot; i >= 2; --i) {//enum circle length for (int j = 1; j <= n; ++j) for (int k = 0; k <= n; ++k) for (int l = 0; l <= n; ++l) f[j][k][l] = -inf; gao(1, 0, i); ans = max(ans, f[1][0][m]); } printf("%.2lf\n", ans); return 0; }