BZOJ1722 [Usaco2006 Mar] Milk Team Select 产奶比赛

直接树形dp就好了恩

令$f[i][j][t]$表示以$i$为根的子树，选出来的点存在$j$对父子关系，$t$表示$i$这个点选或者没选，的最大产奶值

分类讨论自己和儿子分别有没有选，然后转移一下就好了。。。恩，详情看代码好了

/**************************************************************
    Problem: 1722
    User: rausen
    Language: C++
    Result: Accepted
    Time:28 ms
    Memory:2808 kb
****************************************************************/
 
#include <cstdio>
#include <algorithm>
 
using namespace std;
const int N = 505;
const int inf = 1e9;
 
struct edge {
    int next, to;
    edge(int _n = 0, int _t = 0): next(_n), to(_t) {}
} e[N];
 
struct tree_node {
    int v, fa;
} tr[N];
 
int n, tar, ans;
int first[N], tot;
int f[N][N][2];
 
inline void add_edge(int x, int y) {
    e[++tot] = edge(first[x], y);
    first[x] = tot;
}
 
#define y e[x].to
void dfs(int p) {
    int t1, t2, tmp, i, j, x;
    static int t[N];
    f[p][0][0] = 0, f[p][0][1] = tr[p].v;
    for (i = 1; i < n; ++i)
        f[p][i][0] = f[p][i][1] = -inf;
    if (first[p] == 0) return;
    for (x = first[p]; x; x = e[x].next) {
        dfs(y);
        for (t1 = 0; t1 < 2; ++t1) {
            for (j = 0; j < n; ++j) t[j] = f[p][j][t1];
            for (t2 = 0; t2 < 2; ++t2) {
                tmp = t1 && t2 && p;
                for (i = 0; i < n; ++i) if (f[y][i][t2] != -inf)
                    for(j = n - 1; i + tmp <= j; --j)
                        if (f[p][j - i - tmp][t1] != -inf)
                            t[j] = max(t[j], f[p][j - i - tmp][t1] + f[y][i][t2]);
            }
            for (j = 0; j < n; ++j) f[p][j][t1] = max(f[p][j][t1], t[j]);
        }
    }
     
}
#undef y
 
int main() {
    int i;
    scanf("%d%d", &n, &tar);
    for (i = 1; i <= n; ++i) {
        scanf("%d%d", &tr[i].v, &tr[i].fa);
        add_edge(tr[i].fa, i);
    }
    dfs(0);
    for (ans = n - 1; ~ans && f[0][ans][0] < tar; --ans);
    printf("%d
", ans);
    return 0;
}