）逼着自己写DP
题意：在一棵树上选出最多的叶子节点，使得叶子节点的值 减去 各个叶子节点到根节点的消耗 >= 0；
思路：
树上分组背包DP，设dp[u][k] 表示 到以u节点为根节点选k个叶子能得到的最大值。则可得到状态转移方程，
dp[u][k] = max(dp[u][k], dp[u][k-t] + dp[v][t] - cost),其中的t表示不同的可能，需要枚举。
#include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <cstdlib> #include <iomanip> #include <bitset> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <stack> #include <cmath> #include <queue> #include <list> #include <map> #include <set> #include <cassert> using namespace std; #define lson (l , mid , rt << 1) #define rson (mid + 1 , r , rt << 1 | 1) #define debug(x) cerr << #x << " = " << x << " "; #define pb push_back #define pq priority_queue typedef long long ll; typedef unsigned long long ull; //typedef __int128 bll; typedef pair<ll ,ll > pll; typedef pair<int ,int > pii; typedef pair<int,pii> p3; //priority_queue<int> q;//这是一个大根堆q //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q #define fi first #define se second //#define endl ' ' #define OKC ios::sync_with_stdio(false);cin.tie(0) #define FT(A,B,C) for(int A=B;A <= C;++A) //用来压行 #define REP(i , j , k) for(int i = j ; i < k ; ++i) #define max3(a,b,c) max(max(a,b), c); #define min3(a,b,c) min(min(a,b), c); //priority_queue<int ,vector<int>, greater<int> >que; const ll mos = 0x7FFFFFFF; //2147483647 const ll nmos = 0x80000000; //-2147483648 const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3f; //18 const int mod = 1e9+7; const double esp = 1e-8; const double PI=acos(-1.0); const double PHI=0.61803399; //黄金分割点 const double tPHI=0.38196601; template<typename T> inline T read(T&x){ x=0;int f=0;char ch=getchar(); while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar(); while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); return x=f?-x:x; } /*-----------------------showtime----------------------*/ const int maxn = 3009; vector<pii>mp[maxn]; int n,m; int dp[maxn][maxn],a[maxn]; int dfs(int u, int fa){ int res = 0; if(mp[u].size() == 0) { dp[u][1] = a[u]; return 1; } for(int i=0; i<mp[u].size(); i++){ int v = mp[u][i].fi; if(v == fa) continue; res += dfs(v, u); dp[u][0] = 0; for(int j=res; j >= 1; j--){ for(int k=1; k<=j; k++){ dp[u][j] = max(dp[u][j], dp[u][j-k] + dp[v][k] - mp[u][i].se); } } } return res; } int main(){ scanf("%d%d", &n, &m); memset(dp, ~inf, sizeof(dp)); for(int i=1; i<=n-m; i++){ int k; scanf("%d", &k); for(int j=1; j<=k; j++) { int v,w; scanf("%d%d", &v, &w); mp[i].pb(pii(v,w)); } } for(int i=n-m+1; i<=n; i++) scanf("%d", &a[i]); int tmp = dfs(1,-1); for(int i=tmp; i>=1; i--) { if(dp[1][i] >= 0) { printf("%d ", i); return 0; } } return 0; }