Codeforces 815C. Karen and Supermarket【树形DP】

LINK

---

思路

首先发现依赖关系是一个树形的结构

然后因为直接算花多少钱来统计贡献不是很好

因为数组开不下

那就可以算一个子树里面选多少个的最小代价就可以了

注意统计贡献的时候用优惠券的答案只能在1号点进行统计

---

//Author: dream_maker
#include<bits/stdc++.h>
using namespace std;
//----------------------------------------------
//typename
typedef 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 typename
const int INF_of_int = 1e9;
const ll INF_of_ll = 1e18;
//fast read and write
template <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;
}