树形DP，一脸蒙蔽。看了题解才发现它转移状态与方程真不愧神题！
(f[x][y])表示(x)的(y)层以下的所有点都已经覆盖完，还需要覆盖上面的(y)层的最小代价。
(g[x][y])表示(x)子树中所有点都已经覆盖完，并且(x)还能向上覆盖(y)层的最小代价。
(g[u][j]=min(g[u][j]+f[v][j],g[v][j+1]+f[u][j+1]))
(f[u][j] = Σf[v][j-1]f[u][j])
(g[u][j] = min(g[u][j], g[u][j+1]))
(f[u][j] = min(f[u][j], f[u][j-1]))
我在口胡些啥
#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("
----------
")
#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 = 500007;
int n, m, D;
struct Edge{
int nxt, pre;
}e[N << 1];
int head[N], cntEdge;
inline void add(int u, int v){
e[++cntEdge] = (Edge){ head[u], v}, head[u] = cntEdge;
}
int val[N];
long long g[N][29], f[N][29];
int vis[N];
inline void DFS(int u, int fa){
if(vis[u]){
f[u][0] = g[u][0] = val[u];
}
R(i,1,D){
g[u][i] = val[u];
}
g[u][D + 1]= 0x3f3f3f3f;
for(register int i = head[u]; i; i = e[i].nxt){
int v = e[i].pre;
if(v == fa) continue;
DFS(v, u);
nR(j,D,0){
g[u][j] = Min(g[u][j] + f[v][j], g[v][j + 1] + f[u][j + 1]);
}
nR(j,D,0){
g[u][j] = Min(g[u][j], g[u][j + 1]);
}
f[u][0] = g[u][0];
R(j,1,D + 1){
f[u][j] += f[v][j - 1];
}
R(j,1,D + 1){
f[u][j] = Min(f[u][j], f[u][j - 1]);
}
}
}
int main(){
//FileOpen();
io >> n >> D;
R(i,1,n){
io >> val[i];
}
io >> m;
R(i,1,m){
int x;
io >> x;
vis[x] = 1;
}
R(i,2,n){
int u, v;
io >> u >> v;
add(u, v);
add(v, u);
}
DFS(1, 0);
printf("%lld
",f[1][0] );
return 0;
}
/*
3 1
1 1 1
3
1 1 1
1 2
1 3
14 2
1 1 1 1 1 1 1 1 1 1 1 1 1 1
14
1 2 3 4 5 6 7 8 9 10 11 12 13 14
1 2
1 3
1 4
2 5
2 6
3 7
3 8
3 9
6 10
10 11
11 12
12 13
12 14
*/
