Codeforces 815C. Karen and supermarket "tree DP"

LINK

Ideas

First, we find that the dependency is a tree-shaped structure.
And then it's not very good to have a direct count of how much it cost to contribute.
Because the array cannot be opened
Then you can count the minimum cost of a subtree.

Pay attention to the statistical contribution of the time with the coupon answer can only be counted at number 1th.

Author:dream_maker#include<bits/stdc++.h>using namespace std;//-----------------------------------------  -----//typenametypedef Long Long ll;//convenient for#define fu (A, B, c) for (int a = B; a <= C; ++a) #define FD (A, B, c) for (Int. a = B; a >= C;--a) #define FV (A, b) for (int a = 0; a < (signed) b.size (); ++a)//inf of different Typenamec  onst int inf_of_int = 1e9;const ll inf_of_ll = 1e18;//fast Read and writetemplate <typename t>void read (T &x) {  bool W = 1;x = 0;  char C = GetChar ();  while (!isdigit (c) && c! = '-') c = GetChar ();  if (c = = '-') w = 0, c = getchar ();    while (IsDigit (c)) {x = (x<<1) + (x<<3) + C-' 0 ';  c = GetChar (); } if (!w) x =-X;}    Template <typename t>void Write (T x) {if (x < 0) {Putchar ('-');   x =-X;  } if (x > 9) Write (X/10); Putchar (x 10 + ' 0 ');} ----------------------------------------------const int N = 5010;struct edge{int v, NXT;} E[n << 1];int head[n], tot = 0;ll c[N], D[n], siz[n], n;ll dp[n][n][2], ans = 0, b;void add (int u, int v) {E[++tot] = (Edge) {V, head[u]}; Head[u] = tot;}  void Dfs (int u, int fa) {Siz[u] = 1;  DP[U][1][1] = C[u]-d[u];  Dp[u][1][0] = C[u];    for (int i = head[u]; i; i = e[i].nxt) {int v = E[I].V;    if (v = = FA) continue;    DFS (v, u); FD (J, Siz[u], 0) fd (k, siz[v], 0) {dp[u][j + k][1] = min (dp[u][j + k][1], dp[u][j][1] + min (dp[v][k][0), dp[v        ][K][1]));        Dp[u][j + k][0] = min (dp[u][j + k][0], dp[u][j][0] + dp[v][k][0]);  } Siz[u] + = Siz[v];    } Fu (i, ans + 1, siz[u]) {if (dp[u][i][0] <= B) ans = i;  else break;  }}int Main () {memset (DP, 0x3f, sizeof (DP)); Read (n);  Read (B); Fu (i, 1, n) {Read (c[i]);    Read (D[i]);    Dp[i][0][0] = 0; if (i > 1) {int u;      Read (U);      Add (i, u);    Add (U, i);  }} DFS (1, 0);  Fu (i, ans + 1, N) if (dp[1][i][1] <= B) ans = i;  Write (ANS); return 0;}

Codeforces 815C. Karen and supermarket "tree DP"