cogs 2652. 秘术「天文密葬法」(0/1分数规划 长链剖分 二分答案 dp

http://cogs.pro:8080/cogs/problem/problem.php?pid=vSXNiVegV

题意：给个树，第i个点有两个权值ai和bi，现在求一条长度为m的路径，使得Σai/Σbi最小。

思路：二分答案得p，把每个点权值变成ai-p*bi，看是否存在长为一条长为m的路使总和<=0。

tag数组表示从当前位置沿最长链走到底的值，dp数组初值表示从当前位置的重儿子走到底的值（加负号），用tag[…]+dp[..]维护从当前节点往下走若干步得到的最小值（只更新dp数组

 # include <stdio.h>
 # include <string.h>
 # include <iostream>
 # include <algorithm>
 // # include <bits/stdc++.h> using namespace std; typedef long long ll;
 typedef long double ld;
 typedef unsigned long long ull;
 const int N = 3e4 + , M = 6e4 + ;
 const int mod = 1e9+; int n, m, A[N * ], B[N * ], head[N], nxt[M], to[M], tot = ;
 inline void add(int u, int v) {
     ++tot; nxt[tot] = head[u]; head[u] = tot; to[tot] = v;
 }
 inline void adde(int u, int v) {
     add(u, v), add(v, u);
 } int dep[N], mxd[N], son[N], sz[N];
 inline void pre_dfs(int x, int fa = ) {
     dep[x] = dep[fa] + ;
     mxd[x] = dep[x]; son[x] = ;
     for (int i=head[x]; i; i=nxt[i]) {
         if(to[i] == fa) continue;
         pre_dfs(to[i], x);
         if(mxd[to[i]] > mxd[x]) mxd[x] = mxd[to[i]], son[x] = to[i];
     }
     sz[x] = mxd[x] - dep[x];
 } int pos[N], idx;
 inline void pre_pos(int x, int fa = ) {
     pos[x] = ++idx;
     if(son[x]) pre_pos(son[x], x);
     for (int i=head[x]; i; i=nxt[i])
         if(to[i] != fa && to[i] != son[x]) pre_pos(to[i], x);
 } double mid_check, ans;
 double dp[N], tag[N];
 inline void solve(int x, int fa = ) {
     double *f = &dp[pos[x]], C = (double)A[x] - mid_check * B[x];
     if(son[x] == ) {    //leaf
         f[] = ; tag[x] = C;
         if(m == ) ans = min(ans, tag[x]);
         return ;
     }
     solve(son[x], x); f[] = -tag[son[x]];
     tag[x] = tag[son[x]] + C;
     for (int i=head[x], y; i; i=nxt[i]) {
         if(to[i] == fa || to[i] == son[x]) continue;
         solve(y = to[i], x);
         double *g = &dp[pos[y]];
         for (int j=; j<=sz[y] && j<m; ++j)
             if(m--j <= sz[x]) ans = min(ans, f[m--j] + tag[x] + g[j] + tag[y]);
          for (int j=; j<=sz[y]; ++j) f[j+] = min(f[j+], g[j] + tag[y] + C - tag[x]);
     }
     if(m <= sz[x]) ans = min(ans, f[m] + tag[x]);
 } inline bool chk(double x) {
      ans = 1e18; mid_check = x;
      solve();
     return ans <= ;
 } int main() {
     freopen("cdcq_b.in", "r", stdin);
     freopen("cdcq_b.out", "w", stdout);
     cin >> n >> m;
     for (int i=; i<=n; ++i) scanf("%d", A+i);
     for (int i=; i<=n; ++i) scanf("%d", B+i);
     if(m == -) {
         double ans = 1e18;
         for (int i=; i<=n; ++i) ans = min(ans, (double)A[i]/B[i]);
         printf("%.2lf\n", ans);
         return ;
     }
     for (int i=, u, v; i<n; ++i) {
         scanf("%d%d", &u, &v);
         adde(u, v);
     }
     --m;
     pre_dfs();
     pre_pos();
     double l = , r = 1e11, mid;
     while(r-l > 1e-) {
         mid = (l+r)/2.0;
         if(chk(mid)) r = mid;
         else l = mid;
     }
     if(l > 5e10) puts("-1");
     else printf("%.2lf\n", l);
     return ;
 }