bzoj 4033: [HAOI2015]树上染色 [树形DP]

4033: [HAOI2015]树上染色

---

我写的可是(O(n^2))的树形背包！

注意j倒着枚举，而k要正着枚举，因为k可能从0开始，会使用自己更新一次

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const int N = 2005, P = 1e9+7;
inline int read() {
    char c=getchar(); int x=0,f=1;
    while(c<'0'||c>'9') {if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9') {x=x*10+c-'0';c=getchar();}
    return x*f;
}

int n, m, mm, u, v;
struct edge{int v, ne, w;} e[N<<1];
int cnt, h[N];
inline void ins(int u, int v, int w) { 
	e[++cnt] = (edge){v, h[u], w}; h[u] = cnt;
	e[++cnt] = (edge){u, h[v], w}; h[v] = cnt;
}

ll f[N][N]; int size[N];
void dp(int u, int fa) { //printf("dp %d %d
", u, fa);
	size[u] = 1;
	for(int i=h[u]; i; i=e[i].ne) {
		int v = e[i].v, w = e[i].w;
		if(v == fa) continue;
		dp(v, u);
		for(int j = min(size[u] + size[v], m); j >= 0; j--) {
			int _ = min(j, size[v]);
			for(int k = max(0, j - size[u]); k <= _; k++)
				f[u][j] = max(f[u][j], f[v][k] + f[u][j-k] + (ll) w * ( k * (m-k) + (size[v] - k) * (mm - size[v] + k) ) );
		}
		size[u] += size[v];
	}
	//printf("look %d  %d
", u, size[u]);
	//for(int i=0; i<=min(size[u], m); i++) printf("f %d %d  %lld
", u, i, f[u][i]);
	//puts("end
");
}
int main() {
	//freopen("in", "r", stdin);
	freopen("haoi2015_t1.in", "r", stdin);
	freopen("haoi2015_t1.out", "w", stdout);
	n = read(); m = read(); mm = n - m;
	for(int i=1; i<n; i++) u = read(), v = read(), ins(u, v, read());
	dp(1, 0);
	printf("%lld
", f[1][m]);
}