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 + 4;
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 = 0, p = 1; char c = getchar();
    for(; c > '9' || c < '0'; c = getchar()) if(c == '-') p = -1;
    for(; c >= '0' && c <= '9'; c = getchar()) X = X * 10 + c - '0';
    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], 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);
    }
    if(u >= n - m + 1) F[u][1] = a[u], sum[u] = 1;
}

int main()
{
    n = rd; m = rd;
    rep(i, 1, n - m) {
        R int cnt = rd;
        rep(j, 1, cnt) {
            R int u = rd, val = rd;
            add(i, u, val);
        }
    }
    rep(i, n - m + 1, n) a[i] = rd;
    rep(i, 1, n) rep(j, 1, m) F[i][j] = -inf;
    dp(1);
    per(i, m, 0) 
        if(F[1][i] >= 0) return printf("%d\n", i), 0;
}