SPOJ FTOUR2 Free tour II

Description

After the success of 2nd anniversary (take a look at problem FTOUR for more details), this 3rd year, Travel Agent SPOJ goes on with another discount tour.

The tour will be held on ICPC island, a miraculous one on the Pacific Ocean. We list N places (indexed from 1 to N) where the visitors can have a trip. Each road connecting them has an interest value, and this value can be negative (if there is nothing interesting to view there). Simply, these N places along with the roads connecting them form a tree structure. We will choose two places as the departure and destination of the tour.

Since September is the festival season of local inhabitants, some places are extremely crowded (we call them crowded places). Therefore, the organizer of the excursion hopes the tour will visit at most K crowded places (too tiring to visit many of them) and of course, the total number of interesting value should be maximum.

Briefly, you are given a map of N places, an integer K, and M id numbers of crowded place. Please help us to find the optimal tour. Note that we can visit each place only once (or our customers easily feel bored), also the departure and destination places don't need to be different.

Input

There is exactly one case. First one line, containing 3 integers N K M, with 1 <= N <= 200000, 0 <= K <= M, 0 <= M <= N.

Next M lines, each line includes an id number of a crowded place.

The last (N - 1) lines describe (N - 1) two-way roads connected N places, form a b i, with a, b is the id of 2 places, and i is its interest value (-10000 <= i <= 10000).

Output

Only one number, the maximum total interest value we can obtain.

Example

Input:
8 2 3
3
5
7
1 3 1
2 3 10
3 4 -2
4 5 -1
5 7 6
5 6 5
4 8 3


Output:
12

Explanation

We choose 2 and 6 as the departure and destination place, so the tour will be 2 -> 3 -> 4 -> 5 -> 6, total interest value = 10 + (-2) + (-1) + 5 = 12

* Added some unofficial cases

Submit solution!

hide comments 

	w703710691d: 2015-09-04 15:55:24  why I always get a re?
	lu: 2015-06-01 17:08:24  the result turned to TLE QAQ
	lu: 2015-06-01 16:03:01  my program passed the following tests but it got WA.I don't know what's wrong
	kid: 2015-01-07 15:17:10  I chanllenged many codes using the following input  2 0 0  1 2 1  pay attention to the initialization
	yzx: 2012-09-23 07:05:00  answer:6
	Exia_cai: 2012-09-14 08:07:12  here is a test：  5 3 4  2  3  4  5  1 2 1  2 3 3  1 4 1  1 5 2
	谢良: 2011-06-09 12:31:58  300
	Zhao Tao: 2010-04-27 03:13:28  The test data of this problem is somewhat weak. I successfully chanllenged a code which was accepted using the following input:  7 5 6  2  3  4  5  6  7  1 7 100  1 5 100  5 6 100  1 2 1  2 3 1  3 4 1

Hint

Added by:	Thanh-Vy Hua
Date:	2007-09-28
Time limit:	0.100s-0.443s
Source limit:	50000B
Memory limit:	1536MB
Cluster:	Cube (Intel G860)
Languages:	All except: ERL JS NODEJS PERL 6 VB.net
Resource:	Adapted from Preslav Le's problem, first used in Bulgarian OI 07

树分治，统计一下每个点出发的经过的点数与距离，我这里选择用树状数组统计前缀的最大值。

#include<queue>
#include<cstdio>
#include<vector>
#include<iostream>
#include<algorithm>
using namespace std;
const int low(int x) { return x&-x; }
const int INF = 0x7FFFFFFF;
const int maxn = 4e5 + 10;
int n, w, m, x, y, z, f[maxn];

struct Tree
{
	int ft[maxn], nt[maxn], u[maxn], v[maxn], sz;
	int mx[maxn], ct[maxn], vis[maxn];
	int d[maxn], D[maxn], t;

	void clear(int n)
	{
		mx[sz = 0] = INF;
		for (int i = 1; i <= n; i++) ft[i] = -1, f[i] = vis[i] = 0;
	}
	void AddEdge(int x, int y, int z)
	{
		u[sz] = y;	v[sz] = z;	nt[sz] = ft[x];	ft[x] = sz++;
		u[sz] = x;	v[sz] = z;	nt[sz] = ft[y]; ft[y] = sz++;
	}
	int dfs(int x, int fa, int sum)
	{
		int y = mx[x] = (ct[x] = 1) - 1;
		for (int i = ft[x]; i != -1; i = nt[i])
		{
			if (vis[u[i]] || u[i] == fa) continue;
			int z = dfs(u[i], x, sum);
			ct[x] += ct[u[i]];
			mx[x] = max(mx[x], ct[u[i]]);
			y = mx[y] < mx[z] ? y : z;
		}
		mx[x] = max(mx[x], sum - ct[x]);
		return mx[x] < mx[y] ? x : y;
	}
	int get(int x, int fa, int k, int dep)
	{
		int y = k;
		d[k] = max(d[k], dep);
		for (int i = ft[x]; i != -1; i = nt[i])
		{
			if (vis[u[i]] || u[i] == fa) continue;
			y = max(get(u[i], x, k + f[u[i]], dep + v[i]), y);
		}
		return y;
	}
	int find(int x)
	{
		t = get(x, -1, f[x], 0) + 1;
		int ans = 0;
		for (int i = 0; i <= t; i++) D[i] = 0;
		for (int i = ft[x]; i != -1; i = nt[i])
		{
			if (vis[u[i]]) continue;
			int len = get(u[i], x, f[x] + f[u[i]], v[i]);
			for (int j = f[x] + f[u[i]]; j <= len; j++) d[j] = -INF;
			get(u[i], x, f[x] + f[u[i]], v[i]);
			for (int j = f[x] + f[u[i]]; j <= len; j++)
			{
				int Max = -INF;
				for (int k = min(w + f[x] - j + 1, t); k > 0; k -= low(k)) Max = max(Max, D[k]);
				if (Max != -INF) ans = max(ans, Max + d[j]);
			}
			for (int j = f[x] + f[u[i]]; j <= len; j++)
			{
				for (int k = j + 1; k <= t; k += low(k)) D[k] = max(D[k], d[j]);
			}
		}
		for (int k = min(w + 1, t); k > 0; k -= low(k)) ans = max(ans, D[k]);
		return ans;
	}
	int work(int x, int sum)
	{
		int y = dfs(x, -1, sum);
		int ans = find(y);	vis[y] = 1;
		for (int i = ft[y]; i != -1; i = nt[i])
		{
			if (vis[u[i]]) continue;
			ans = max(ans, work(u[i], ct[u[i]] > ct[y] ? sum - ct[y] : ct[u[i]]));
		}
		return ans;
	}
}solve;

int main()
{
	while (scanf("%d%d%d", &n, &w, &m) != EOF)
	{
		solve.clear(n);
		while (m--) scanf("%d", &x), f[x] = 1;
		for (int i = 1; i < n; i++)
		{
			scanf("%d%d%d", &x, &y, &z);
			solve.AddEdge(x, y, z);
		}
		printf("%d\n", solve.work(1, n));
	}
	return 0;
}