1825. Free tour IIProblem code: FTOUR2 |
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 atree 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 themcrowded 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
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <iomanip>
#include <cassert>
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define ff(i, n) for(int i=0;i<(n);i++)
#define fff(i, n, m) for(int i=(n);i<=(m);i++)
#define dff(i, n, m) for(int i=(n);i>=(m);i--)
#define travel(e, u) for(int e = u, v = vv[u]; e; e = nxt[e], v = vv[e])
#define Bit(n) (1LL<<(n))
#define And(a, b) ((a) & (b))
#define Xor(a, b) ((a) ^ (b))
#define Or(a, b) ((a) | (b))
typedef long long LL;
typedef unsigned long long ULL;
void work();
int main()
{
#ifdef ACM
freopen("in.txt", "r", stdin);
#endif // ACM
work();
}
void scanf(int & x, char c = 0)
{
while((c = getchar()) < '0' || c > '9');
x = c - '0';
while((c = getchar()) >= '0' && c <= '9') x = x * 10 + (c - '0');
}
#define rep(i,a,b) for(int i = (a);i<(b);++i)
#define rrep(i,a,b) for(int i = (a);i>=(b);--i)
#define clr(a,x) memset(a,x,sizeof(a))
/*****************************************華麗分割線**********************************************/
const int maxv = 222222;
/// Global
int n, m, k;
int ans;
/// Edge
int edge[maxv], ecnt;
int nxt[maxv * 2], vv[maxv * 2], ww[maxv * 2];
/// Tree
bool blk[maxv], vis[maxv];
int siz[maxv], mon[maxv], dad[maxv];
int root, minN, maxD;
/// Tree Node
int all[maxv], allK;
int now[maxv], nowK;
struct Node
{
int v, w, d;
Node(int v = 0, int w = 0, int d = 0) : v(v), w(w), d(d) {}
bool operator<(const Node & cmp) const
{
return d < cmp.d;
}
};
void init()
{
/// Edge initial
clr(edge, 0);
ecnt = 2;
/// Ans Initial
ans = 0;
/// Tree Initial
clr(blk, 0);
clr(vis, 0);
}
void addEdge(int u, int v, int w, int first[])
{
nxt[ecnt] = first[u], vv[ecnt] = v, ww[ecnt] = w, first[u] = ecnt++;
}
void input()
{
ff(i, m)
{
int u;
scanf("%d", &u);
blk[u] = true;
}
ff(i, n-1)
{
int u, v, w;
scanf("%d%d%d", &u, &v, &w);
addEdge(u, v, w, edge);
addEdge(v, u, w, edge);
}
}
void dfsSize(int u)
{
siz[u] = 1;
mon[u] = 0;
travel(e, edge[u]) if(!vis[v] && v != dad[u])
dad[v] = u, dfsSize(v), siz[u] += siz[v], mon[u] = max(mon[u], siz[v]);
}
int r;
void dfsRoot(int u)
{
mon[u] = max(mon[u], siz[r] - siz[u]);
if(mon[u] < minN) minN = mon[u], root = u;
travel(e, edge[u]) if(!vis[v] && v != dad[u])
dfsRoot(v);
}
int bb;
void dfsDeep(int u)
{
if(blk[u]) bb++;
maxD = max(maxD, bb);
travel(e, edge[u]) if(!vis[v] && v != dad[u])
dad[v] = u, dfsDeep(v);
if(blk[u]) bb--;
}
int wv;
void dfsValue(int u)
{
if(blk[u]) bb++;
now[bb] = max(now[bb], wv);
travel(e, edge[u]) if(!vis[v] && v != dad[u])
wv += ww[e], dfsValue(v), wv -= ww[e];
if(blk[u]) bb--;
}
void dfs(int u)
{
/// Get root
dad[u] = 0, dfsSize(u);
r = u, minN = n, dfsRoot(u);
/// Get son and sort
u = root;
vector<Node> son;
travel(e, edge[u]) if(!vis[v])
bb = 0, dad[v] = u, maxD = 0, dfsDeep(v), son.push_back(Node(v, ww[e], maxD));
sort(son.begin(), son.end());
if(son.size() == 0) return;
allK = -1;
int kk = k;
if(blk[u]) kk--;
if(kk >= 0)
ff(i, son.size())
{
nowK = son[i].d;
fff(j, 0, nowK) now[j] = Bit(31);
bb = 0;
wv = son[i].w;
dfsValue(son[i].v);
fff(j, 1, nowK) now[j] = max(now[j], now[j-1]);
ans = max(ans, now[min(kk, nowK)]);
if(allK == -1)
{
}
else if(allK < kk)
{
if(allK + nowK <= kk)
ans = max(ans, now[nowK] + all[allK]);
else
fff(j, max(0, kk-nowK), allK)
ans = max(ans, all[j] + now[kk-j]);
}
else
{
dff(j, kk, 0)
ans = max(ans, all[j] + now[kk-j]);
}
/// Merge
fff(j, 0, allK) all[j] = max(all[j], now[j]);
if(allK >= 0) all[0] = max(all[0], now[0]);
fff(j, max(allK, 0) + 1, nowK) all[j] = max(now[j], all[j-1]);
allK = nowK;
}
vis[u] = true;
travel(e, edge[u]) if(!vis[v])
dfs(v);
}
void work()
{
while(scanf("%d%d%d", &n, &k, &m) == 3)
{
init();
input();
dfs(1);
printf("%d\n", ans);
}
}