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
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<cstring> 2 #include<cstdio> 3 #include<algorithm> 4 #define rd read() 5 #define R register 6 #define rep(i,a,b) for(register int i = (a); i <= (b); ++i) 7 #define per(i,a,b) for(register int i = (a); i >= (b); --i) 8 using namespace std; 9 10 const int N = 3e3 + 4; 11 const int inf = 1e8; 12 13 int n, m, tot, head[N]; 14 int a[N], sum[N], F[N][N]; 15 16 struct edge { 17 int nxt, to, val; 18 }e[N]; 19 20 inline int read() { 21 int X = 0, p = 1; char c = getchar(); 22 for(; c > '9' || c < '0'; c = getchar()) if(c == '-') p = -1; 23 for(; c >= '0' && c <= '9'; c = getchar()) X = X * 10 + c - '0'; 24 return X * p; 25 } 26 27 inline void add(R int u, R int v, R int val) { 28 e[++tot].to = v; 29 e[tot].nxt = head[u]; 30 e[tot].val = val; 31 head[u] = tot; 32 } 33 34 inline int cmax(int A, int B) { 35 return A > B ? A : B; 36 } 37 38 inline void dp(R int u) { 39 for(R int i = head[u]; i; i = e[i].nxt) { 40 R int nt = e[i].to; 41 dp(nt); 42 sum[u] += sum[nt]; 43 per(k, sum[u], 0) per(j, sum[nt], 1) if(k >= j)F[u][k] = cmax(F[u][k], F[u][k - j] + F[nt][j] - e[i].val); 44 } 45 if(u >= n - m + 1) F[u][1] = a[u], sum[u] = 1; 46 } 47 48 int main() 49 { 50 n = rd; m = rd; 51 rep(i, 1, n - m) { 52 R int cnt = rd; 53 rep(j, 1, cnt) { 54 R int u = rd, val = rd; 55 add(i, u, val); 56 } 57 } 58 rep(i, n - m + 1, n) a[i] = rd; 59 rep(i, 1, n) rep(j, 1, m) F[i][j] = -inf; 60 dp(1); 61 per(i, m, 0) 62 if(F[1][i] >= 0) return printf("%d\n", i), 0; 63 }