[SP1825] Free tour II

/*
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[题解]
https://www.luogu.org/blog/IRving1-1/solution-sp1825
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O(Nlog^2)做法,vjudge上写的是时限100ms,过2e5数据
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统计tmp[i]为有i个黑点的最长路径,进行转移
合并的顺序很巧妙,也很重要,这里倒序枚举当前子树的j(tmp[j]),则可以做到控制维护之前子树cur(maxn[cur])单调递增
用maxn[i]记录小于等于i个黑点的最长路径,更新答案完了以后用当前的tmp[]更新maxn[]
记得清空tmp[]和maxn[]
-----------------------2019.2.12
*/
#pragma GCC optimize(2)
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<set>
#include<vector>
#include<map>
#include<queue>
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define enter putchar('\n')
#define fr friend inline
#define y1 poj
#define mp make_pair
#define pr pair<int,int>
#define fi first
#define sc second
#define pb push_back
#define lowbit(x) x & (-x)
#define B printf("Bug\n");

using namespace std;
typedef long long ll;
const int M = 200005;
const int N = 2000005;
const int INF = 1e9;
const double eps = 1e-7;

int read()
{
    int x = 0,op = 1;char ch = getchar();
    while(ch < '0' || ch > '9') {if(ch == '-') op = -1;ch = getchar();}
    while(ch >= '0' && ch <= '9') x = x * 10 + ch - '0',ch = getchar();
    return x * op;
}

struct edge
{
   int next,to,from,v;
}e[M<<1];

int n,k,m,maxn[M],G,size[M],dis[M],dep[M],sum,x,y,z,head[M],ecnt,hson[M],ans,tmp[M],mdep;
bool black[M],vis[M];
vector <pr> v;

void add(int x,int y,int z)
{
   e[++ecnt].to = y;
   e[ecnt].next = head[x];
   e[ecnt].from = x;
   e[ecnt].v = z;
   head[x] = ecnt;
}

void getG(int x,int fa)
{
   size[x] = 1,hson[x] = 0;
   for(int i = head[x];i;i = e[i].next)
   {
      if(e[i].to == fa || vis[e[i].to]) continue;
      getG(e[i].to,x);
      size[x] += size[e[i].to],hson[x] = max(hson[x],size[e[i].to]);
   }
   hson[x] = max(hson[x],sum - size[x]);
   if(hson[x] < hson[G]) G = x;
}

void getdis(int x,int fa,int d,int depth)
{
   dis[x] = d,dep[x] = depth,mdep = max(mdep,dep[x]);
   for(int i = head[x];i;i = e[i].next)
   {
      if(e[i].to == fa || vis[e[i].to]) continue;
      getdis(e[i].to,x,d + e[i].v,depth + black[e[i].to]);
   }
}

void getmaxn(int x,int fa)
{
   tmp[dep[x]] = max(tmp[dep[x]],dis[x]);
   for(int i = head[x];i;i = e[i].next)
   {
      if(vis[e[i].to] || e[i].to == fa) continue;
      getmaxn(e[i].to,x);
   }
}

void solve(int x)
{
   vis[x] = 1,v.clear();
   if(black[x]) k--;
   for(int i = head[x];i;i = e[i].next)
   {
      if(vis[e[i].to]) continue;
      mdep = 0,getdis(e[i].to,x,e[i].v,black[e[i].to]);//算出最长路径有几个黑点mdep
      v.pb(mp(mdep,e[i].to));
   }
   sort(v.begin(),v.end());
   rep(i,0,(int)(v.size()-1))
   {
      getmaxn(v[i].sc,x);//算出有i个黑点的最长路径长度tmp[i]
      int cur = 0;
      if(i != 0)
      per(j,v[i].fi,0)//启发式合并:[倒序]枚举j并控制cur+j<k,这个思想很巧妙,很重要  O(N)
      {
                              //一直没看到这个-1
           while(cur + j < k && cur < v[i-1].fi) cur++,maxn[cur] = max(maxn[cur],maxn[cur-1]);//小于等于i个黑点的最长路径maxn[i]
           if(cur + j <= k) ans = max(ans,maxn[cur] + tmp[j]);
      }
      if(i != (int)(v.size() - 1)) rep(j,0,v[i].fi) maxn[j] = max(maxn[j],tmp[j]),tmp[j] = 0;//小于等于i个黑点的最长路径maxn[i],并准备转移
      else rep(j,0,v[i].fi){if(j <= k) ans = max(ans,max(tmp[j],maxn[j]));tmp[j] = maxn[j] = 0;}//最后一个子树,直接统计
   }
   if(black[x]) k++;
   for(int i = head[x];i;i = e[i].next)
   {
      if(vis[e[i].to]) continue;
      sum = size[e[i].to],G = 0;
      getG(e[i].to,x),solve(G);
   }
}

int main()
{
   n = read(),k = read(),m = read();
   rep(i,1,m) x = read(),black[x] = 1;
   rep(i,1,n-1) x = read(),y = read(),z = read(),add(x,y,z),add(y,x,z);
   sum = n,hson[G] = INF,getG(1,0);
   solve(G);
   printf("%d\n",ans);
   return 0;
}