1722: [Usaco2006 Mar] Milk Team Select 产奶比赛

1722: [Usaco2006 Mar] Milk Team Select 产奶比赛

https://www.lydsy.com/JudgeOnline/problem.php?id=1722

分析：

　　f[u][i][0/1]表示子树u中，有i对相邻的点，最大和是多少。

代码：

 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<cmath>
 #include<iostream>
 #include<cctype>
 #include<set>
 #include<vector>
 #include<queue>
 #include<map>
 #define fi(s) freopen(s,"r",stdin);
 #define fo(s) freopen(s,"w",stdout);
 using namespace std;
 typedef long long LL;

 inline int read() {
     int x=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;
     for(;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;
 }

 const int N = ;
 const int INF = 1e9;

 int f[N][N][], g[N][];
 int head[N], nxt[N], to[N], siz[N], w[N], fa[N];
 int n, x, En;

 void add_edge(int u,int v) {
     ++En; to[En] = v; nxt[En] = head[u]; head[u] = En;
 }

 void dfs(int u) {
     siz[u] = ;
     f[u][][] = w[u];
     f[u][][] = ;
     for (int i=head[u]; i; i=nxt[i]) {
         int v = to[i];
         dfs(v);
         siz[u] += siz[v];
         for (int j=; j<=siz[u]-; ++j) // 相邻的
             for (int k=,lim=min(j, siz[v]-); k<=lim; ++k) { // 子树中相邻的
                 g[j][] = max(g[j][], f[u][j - k][] + max(f[v][k][], f[v][k][]));//cerr << g[j][0] << "\n";
                 if (j - k >= ) g[j][] = max(g[j][], f[u][j - k][] + f[v][k][]); //cerr << g[j][1] << "\n";;
                 if (j - k -  >= ) g[j][] = max(g[j][], f[u][j - k - ][] + f[v][k][]);// cerr << g[j][1] << "\n";;

             }
         for (int j=n; j>=; --j)
             f[u][j][] = g[j][], f[u][j][] = g[j][], g[j][] = g[j][] = -INF;
     }
 }
 int main() {
     n = read(), x = read();
     memset(f, -0x3f, sizeof(f));
     memset(g, -0x3f, sizeof(g)); // 把g[0]初始化了！！！
 //    for (int i=1; i<=n; ++i) g[i][0] = g[i][1] = -INF;
     for (int i=; i<=n; ++i) {
         w[i] = read(), fa[i] = read();
         add_edge(fa[i], i);
     }
     dfs();
     for (int i=n; i>=; --i) {
         if (f[][i][] >= x) { cout << i; return ; }
     }
     cout << -;
     return ;
 }