Luogu 1273 有线电视网 - 树形背包

Description

树形背包， 遍历到一个节点， 枚举它的每个子节点要选择多少个用户进行转移。

Code

 #include<cstring>
 #include<cstdio>
 #include<algorithm>
 #define rd read()
 #define R register
 #define rep(i,a,b) for(register int i = (a); i <= (b); ++i)
 #define per(i,a,b) for(register int i = (a); i >= (b); --i)
 using namespace std;

 const int N = 3e3 + ;
 const int inf = 1e8;

 int n, m, tot, head[N];
 int a[N], sum[N], F[N][N];

 struct edge {
     int nxt, to, val;
 }e[N];

 inline int read() {
     int X = , p = ; char c = getchar();
     for(; c > '' || c < ''; c = getchar()) if(c == '-') p = -;
     for(; c >= '' && c <= ''; c = getchar()) X = X *  + c - '';
     return X * p;
 }

 inline void add(R int u, R int v, R int val) {
     e[++tot].to = v;
     e[tot].nxt = head[u];
     e[tot].val = val;
     head[u] = tot;
 }

 inline int cmax(int A, int B) {
     return A > B ? A : B;
 }

 inline void dp(R int u) {
     for(R int i = head[u]; i; i = e[i].nxt) {
         R int nt = e[i].to;
         dp(nt);
         sum[u] += sum[nt];
         per(k, sum[u], ) per(j, sum[nt], ) if(k >= j)F[u][k] = cmax(F[u][k], F[u][k - j] + F[nt][j] - e[i].val);
     }
     if(u >= n - m + ) F[u][] = a[u], sum[u] = ;
 }

 int main()
 {
     n = rd; m = rd;
     rep(i, , n - m) {
         R int cnt = rd;
         rep(j, , cnt) {
             R int u = rd, val = rd;
             add(i, u, val);
         }
     }
     rep(i, n - m + , n) a[i] = rd;
     rep(i, , n) rep(j, , m) F[i][j] = -inf;
     dp();
     per(i, m, )
         if(F[][i] >= ) return printf("%d\n", i), ;
 }