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，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,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;LL;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<int> 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;for (int i = tot; i >= 2; –i) {//enum circle lengthfor (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;}
愚者用肉体监视心灵，智者用心灵监视肉体
