[Dynamic Route] keys to freedom

The key to freedom (key. PAS/C/CPP) is stored in N rooms, which are connected by n-1 corridor. However, each room has a special protection magic. Under its role, I cannot pass through this room or get the key. Although I can destroy the magic in the room by consuming energy, my energy is limited. So, if I stand in Room 1 first (the protection magic of Room 1 is still valid, that is, if I do not consume energy, I cannot pass Room 1, can't get the key in the room). If my energy is P, how many keys can I get at most? The first line of the input data contains two non-negative integers. The first is N, and the second is P. Next n rows, by 1 ~ The Order N describes each room. Line I + 1 contains two non-negative integers cost and keys, respectively, which are the energy required to cancel the magic in room I and the number of keys in the room. In the next n-1 line, two non-negative integers x and y in each row indicate that room X is connected to room y. An integer in a row of output data indicates that the maximum value of the key is obtained. Sample input: Key. in 5 5 1 2 1 1 1 1 2 3 3 3 4 1 1 3 2 2 4 2 5 output: Key. out 7 data range for 20% of test data, n <= 20 for 30% of test data, n <= 30 for all test data, P, n <= 100, cost <= maxint, keys <= maxint

A signature-based DP.
First, use dual-cross and then DP.
Using keys: F [I] [J] indicates the maximum number of keys that can be obtained using J energy.
Moving equations: f [I] [J] = max (F [son [I] [k] + F [bro [I] [J-K-C [I] + V [I], f [bro [I] [J]).
When loading data, the data is first converted to an undirected sequence, and then the root sequence is not set to a sequence. Otherwise, the data is very heavy; the energy consumed by a certain vertex can be 0, so J = 0 cannot be regarded as a condition.
Accode:

#include <cstdio>#include <cstring>#include <cstdlib>#include <bitset>using std::max;const char fi[] = "key.in";const char fo[] = "key.out";const int maxN = 110;const int maxM = 110;const int MAX = 0x3fffff00;const int MIN = -MAX;struct Edge {int dest; Edge *next; };Edge *edge[maxN];int f[maxN][maxM];int bro[maxN];int son[maxN];int c[maxN];int v[maxN];int n, m;  void init_file()  {    freopen(fi, "r", stdin);    freopen(fo, "w", stdout);  }    void insert(int u, int v)  {    Edge *p = new Edge;    p -> dest = v;    p -> next = edge[u];    edge[u] = p;  }    void readdata()  {    scanf("%d%d", &n, &m);    for (int i = 1; i < n + 1; ++i)      scanf("%d%d", c + i, v + i);    for (int i = 1; i < n; ++i)    {      int u, v;      scanf("%d%d", &u, &v);      insert(u, v);      insert(v, u);    }  }    int DP(int i, int j)  {    if (i == 0) return 0;    if (f[i][j]) return f[i][j];    if (bro[i])      f[i][j] = max(f[i][j], DP(bro[i], j));    if (j >= c[i])    for (int k = 0; k + c[i] < j + 1; ++k)    {      f[i][j] = max(f[i][j], DP(son[i], k)        + DP(bro[i], j - k - c[i]) + v[i]);    }    return f[i][j];  }    void Build(int u, int Last)  {    for (Edge *p = edge[u]; p; p = p -> next)    {      int v = p -> dest;      if (v != Last)      {        Build(v, u);        bro[v] = son[u];        son[u] = v;      }    }  }  void work()  {    Build(1, 0);    DP(1, m);    printf("%d", f[1][m]);  }  int main(){  init_file();  readdata();  work();  exit(0);}