Luogu3177 [HAOI2015]树上染色 （树形DP）

考场上打出来个\(2^n n^2 \log (n)\)，还文件错误RE了。。。
其实这不就是个变了一点点的树形背包，状态是节点\(u\)子树的\(贡献\)。

//#include <iostream>
#include <cstdio>
#include <cstring>
//#include <algorithm>
//#include <cmath>
#define R(a,b,c) for(register int  a = (b); a <= (c); ++ a)
#define nR(a,b,c) for(register int  a = (b); a >= (c); -- a)
#define Max(a,b) ((a) > (b) ? (a) : (b))
#define Min(a,b) ((a) < (b) ? (a) : (b))
#define Fill(a,b) memset(a, b, sizeof(a))
#define Abs(a) ((a) < 0 ? -(a) : (a))
#define Swap(a,b) a^=b^=a^=b
#define ll long long

#define ON_DEBUG

#ifdef ON_DEBUG

#define D_e_Line printf("\n\n----------\n\n")
#define D_e(x)  cout << #x << " = " << x << endl
#define Pause() system("pause")
#define FileOpen() freopen("in.txt","r",stdin);

#else

#define D_e_Line ;
#define D_e(x)  ;
#define Pause() ;
#define FileOpen() ;

#endif

struct ios{
    template<typename ATP>ios& operator >> (ATP &x){
        x = 0; int f = 1; char c;
        for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-')  f = -1;
        while(c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();
        x*= f;
        return *this;
    }
}io;
using namespace std;

const int N = 2007;

#define int long long
struct Edge{
	int nxt, pre, w;
}e[N << 1];
int head[N], cntEdge;
inline void add(int u, int v, int w){
	e[++cntEdge] = (Edge){ head[u], v, w}, head[u] = cntEdge;
}

int n, K;
int f[N][N], siz[N];
inline void DFS(int u, int fa){
	siz[u] = 1;
	f[u][0] = f[u][1] = 0;
	for(register int i = head[u]; i; i = e[i].nxt){
		int v = e[i].pre;
		if(v == fa) continue;
		DFS(v, u);
		siz[u] += siz[v];
		nR(j,Min(siz[u], K),0){
			if(f[u][j] != -1)
				f[u][j] += f[v][0] + siz[v] * (n - K - siz[v]) * e[i].w;
			nR(t,Min(siz[v], j),1){
				if(f[u][j - t] == -1) continue;
				int val = (t * (K - t) + (siz[v] - t) * (n - K - siz[v] + t)) * e[i].w;
				f[u][j] = Max(f[u][j], f[u][j - t] + f[v][t] + val);
			}
		}
	}
}
#undef int
int main(){
#define int long long
//FileOpen();
	io >> n >> K;
	R(i,2,n){
		int u, v, w;
		io >> u >> v >> w;
		add(u, v, w);
		add(v, u, w);
	}
	Fill(f, -1);
	if(n - K < K) K = n - K;
	DFS(1, 0);
	printf("%lld", f[1][K]);
	return 0;
}