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：
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 #include<cmath> 5 #include<iostream> 6 #include<cctype> 7 #include<set> 8 #include<vector> 9 #include<queue> 10 #include<map> 11 #define fi(s) freopen(s,"r",stdin); 12 #define fo(s) freopen(s,"w",stdout); 13 using namespace std; 14 typedef long long LL; 15 16 inline int read() { 17 int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1; 18 for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f; 19 } 20 21 const int N = 1005; 22 const int INF = 1e9; 23 24 int f[N][N][2], g[N][2]; 25 int head[N], nxt[N], to[N], siz[N], w[N], fa[N]; 26 int n, x, En; 27 28 void add_edge(int u,int v) { 29 ++En; to[En] = v; nxt[En] = head[u]; head[u] = En; 30 } 31 32 void dfs(int u) { 33 siz[u] = 1; 34 f[u][0][1] = w[u]; 35 f[u][0][0] = 0; 36 for (int i=head[u]; i; i=nxt[i]) { 37 int v = to[i]; 38 dfs(v); 39 siz[u] += siz[v]; 40 for (int j=0; j<=siz[u]-1; ++j) // 相邻的 41 for (int k=0,lim=min(j, siz[v]-1); k<=lim; ++k) { // 子树中相邻的 42 g[j][0] = max(g[j][0], f[u][j - k][0] + max(f[v][k][1], f[v][k][0]));//cerr << g[j][0] << " "; 43 if (j - k >= 0) g[j][1] = max(g[j][1], f[u][j - k][1] + f[v][k][0]); //cerr << g[j][1] << " ";; 44 if (j - k - 1 >= 0) g[j][1] = max(g[j][1], f[u][j - k - 1][1] + f[v][k][1]);// cerr << g[j][1] << " ";; 45 46 } 47 for (int j=n; j>=0; --j) 48 f[u][j][0] = g[j][0], f[u][j][1] = g[j][1], g[j][0] = g[j][1] = -INF; 49 } 50 } 51 int main() { 52 n = read(), x = read(); 53 memset(f, -0x3f, sizeof(f)); 54 memset(g, -0x3f, sizeof(g)); // 把g[0]初始化了！！！ 55 // for (int i=1; i<=n; ++i) g[i][0] = g[i][1] = -INF; 56 for (int i=1; i<=n; ++i) { 57 w[i] = read(), fa[i] = read(); 58 add_edge(fa[i], i); 59 } 60 dfs(0); 61 for (int i=n; i>=0; --i) { 62 if (f[0][i][0] >= x) { cout << i; return 0; } 63 } 64 cout << -1; 65 return 0; 66 }