题意
luogu的翻译
给定一棵n个点的树，树上有m个黑点，求出一条路径，使得这条路径经过的黑点数小于等于k，且路径长度最大
Sol
点分治辣
如果是等于\(k\)的话，开个桶取\(max\)就好了
而小于等于\(k\)，就可以把桶换成树状数组，求前缀\(max\)
很慢能过
# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(2e5 + 5);
IL int Input(){
RG int x = 0, z = 1; RG char c = getchar();
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}
int n, m, k, mx[_], root, size[_], vis[_], tot, first[_], cnt;
struct Edge{
int to, w, next;
} edge[_ << 1];
int ans, black[_], pass[_], dis[_], t[_], S[_], tk, G[_];
IL void Add(RG int u, RG int v, RG int w){
edge[cnt] = (Edge){v, w, first[u]}, first[u] = cnt++;
}
IL void GetRoot(RG int u, RG int ff){
size[u] = 1, mx[u] = 0;
for(RG int e = first[u]; e != -1; e = edge[e].next){
RG int v = edge[e].to;
if(vis[v] || v == ff) continue;
GetRoot(v, u);
size[u] += size[v];
mx[u] = max(mx[u], size[v]);
}
mx[u] = max(mx[u], tot - size[u]);
if(mx[u] < mx[root]) root = u;
}
IL void GetDeep(RG int u, RG int ff, RG int dd, RG int pp){
if((pp += black[u]) > tk) return;
pass[u] = pp, dis[u] = dd, S[++S[0]] = u, G[++G[0]] = u;
for(RG int e = first[u]; e != -1; e = edge[e].next){
RG int v = edge[e].to, w = edge[e].w;
if(vis[v] || v == ff) continue;
GetDeep(v, u, dd + w, pp);
}
}
IL void Cls(RG int x){
if(!x) t[x] = 0;
for(; x && x <= m; x += x & -x) t[x] = 0;
}
IL void Modify(RG int x, RG int v){
if(!x) t[x] = max(t[x], v);
for(; x && x <= m; x += x & -x) t[x] = max(t[x], v);
}
IL int Query(RG int x){
RG int ret = t[0];
for(; x; x -= x & -x) ret = max(ret, t[x]);
return ret;
}
IL void Calc(RG int u){
for(RG int e = first[u]; e != -1; e = edge[e].next){
RG int v = edge[e].to, w = edge[e].w;
if(vis[v]) continue;
GetDeep(v, u, w, 0);
for(RG int i = 1; i <= S[0]; ++i){
RG int dd = dis[S[i]], pp = tk - pass[S[i]], tt = Query(pp);
ans = max(ans, dd + tt);
}
for(; S[0]; --S[0]){
RG int dd = dis[S[S[0]]], pp = pass[S[S[0]]];
Modify(pp, dd);
}
}
for(; G[0]; --G[0]) Cls(pass[G[G[0]]]);
}
IL void Solve(RG int u){
vis[u] = 1, tk = k - black[u];
if(tk >= 0) Calc(u);
for(RG int e = first[u]; e != -1; e = edge[e].next){
RG int v = edge[e].to;
if(vis[v]) continue;
root = 0, tot = size[v];
GetRoot(v, u), Solve(root);
}
}
int main(RG int argc, RG char* argv[]){
tot = n = Input(), k = Input(), m = Input();
Fill(first, -1), ans = 0;
for(RG int i = 1; i <= m; ++i) black[Input()] = 1;
for(RG int i = 1; i < n; ++i){
RG int u = Input(), v = Input(), w = Input();
Add(u, v, w), Add(v, u, w);
}
mx[0] = n + 1, GetRoot(1, 0), Solve(root);
printf("%d\n", ans);
return 0;
}