Ticket to Ride


Problem Statement :


Simon received the board game Ticket to Ride as a birthday present. After playing it with his friends, he decides to come up with a strategy for the game.

There are n cities on the map and n - 1 road plans. Each road plan consists of the following:

Two cities which can be directly connected by a road.
The length of the proposed road.
The entire road plan is designed in such a way that if one builds all the roads, it will be possible to travel between any pair of cities.

A ticket enables you to travel between two different cities. There are m tickets, and each ticket has a cost associated with it. A ticket is considered to be useful if there is a path between those cities.

Simon wants to choose two cities, u and v, and build a minimal number of roads so that they form a simple path between them. Let St  be the sum of costs of all useful tickets and Sr  be the sum of lengths of all the roads Simon builds. The profit for pair ( u, v )  is defined as St - Sr. Note that u and v are not necessarily unique and may be the same cities.

Given n road plans and m ticket prices, help Simon by printing the value of his maximum possible profit on a new line.


Input Format

The first line contains single positive integer, n , denoting the number of cities.
Each of the n - 1  subsequent lines contains three space-separated integers describing the respective values of u , v, and  l for a road plan, where , 1 <= u ,v <= l and u =/ v . Here, u and v are two cities that the road plan proposes to connect and l  is the length of the proposed road.
The next line contains a single positive integer, m, denoting the number of tickets.
Each of the m subsequent lines contains three space-separated integers describing the respective values of u, v, and c for a ticket from city u  to city v (where c is the cost of the ticket).

Constraints

1  <=  n  <=  2 x 10^5
1  <=  m  <=  10^5
1  <=  l , c  <=  10^9


Output Format

Print a single integer denoting the the maximum profit Simon can make.


Solution :



title-img 


                            Solution in C :

In   C++  :







#include <bits/stdc++.h>

using namespace std;

#define sz(x) ((int) (x).size())
#define forn(i,n) for (int i = 0; i < int(n); ++i)
#define forab(i,a,b)for(int i =int(a);i<int(b);++i)

typedef long long ll;
typedef long double ld;

const int INF = 1000001000;
const ll INFL = 2000000000000001000;
int solve();


int main()
{
srand(2317);
cout.precision(10);
cout.setf(ios::fixed);
#ifdef LOCAL
assert(freopen("test.in", "r", stdin));
#else
#endif
int tn = 1;
for (int i = 0; i < tn; ++i)
solve();
#ifdef LOCAL
cerr << "Time: " << double(clock()) / 
CLOCKS_PER_SEC << '\n';
#endif
}

const int maxn = 500001;
const int maxm = 100001;
const int maxh = 19;

int ROOT;

int n;
vector<pair<int, int>> g[maxn];
int in[maxn], out[maxn];
int timer = 0;

const int base = 1 << maxh;
ll t[base * 2];
ll upd[base * 2];

ll get()
{
return t[1] + upd[1];
}

int l, r, delta;
inline void put(int v = 1, 
int cl = 0, int cr = base)
{
if (r <= cl || cr <= l)
return;
if (l <= cl && cr <= r)
{
upd[v] += delta;
return;
}
int cc = (cl + cr) >> 1;
put(v << 1, cl, cc);
put((v << 1) + 1, cc, cr);
t[v] = max(t[v << 1] + upd[v << 1], 
t[(v << 1) + 1] + upd[(v << 1) + 1]);
}

inline void add(int v, ll delta)
{
l = in[v], r = out[v], ::delta = delta;
put();
}

vector<pair<int, int>> up[maxn];
vector<pair<int, int>> down[maxn];
vector<pair<int, int>> other[maxn];

inline bool is_prev(int u, int v)
{
return in[u] <= in[v] && out[v] <= out[u];
}

ll best = 0;

void go(int u, int prev = ROOT)
{
for (auto p: other[u])
add(p.first, p.second);
for (auto p: down[u])
{
add(p.first, -p.second);
upd[1] += p.second;
}
for (auto p: up[u])
add(p.first, -p.second);
best = max(best, get());
for (auto p: g[u])
{
if (p.first == prev)
continue;
add(p.first, 2 * p.second);
upd[1] -= p.second;
go(p.first, u);
add(p.first, -2 * p.second);
upd[1] += p.second;
}
for (auto p: other[u])
add(p.first, -p.second);
for (auto p: down[u])
{
add(p.first, p.second);
upd[1] -= p.second;
}
for (auto p: up[u])
add(p.first, p.second);
}

pair<int, int> tickets[maxm];
int ticket_cost[maxm];
int visits[maxm];
int needh[maxm];
int step[maxm];
vector<int> endings[maxn];

int st[maxn], sc = 0;

void dfs(int u, int prev = ROOT, ll depth = 0)
{
st[sc++] = u;
for (int id: endings[u])
{
visits[id]++;
if (visits[id] == 1)
needh[id] = sc;
else if (visits[id] == 2)
step[id] = st[needh[id]];
}
in[u] = timer++;
t[in[u] + base] = -depth;

for (auto p: g[u])
if (p.first != prev)
dfs(p.first, u, depth + p.second);

for (int id: endings[u])
visits[id]--;
out[u] = timer;
--sc;
}

int solve()
{
scanf("%d", &n);
ROOT = rand() % n;
forn (i, n - 1)
{
int u, v, l;
scanf("%d %d %d", &u, &v, &l);
--u, --v;
g[u].emplace_back(v, l);
g[v].emplace_back(u, l);
}
int m;
scanf("%d", &m);
forn (i, m)
{
int u, v, c;
scanf("%d %d %d", &u, &v, &c);
--u, --v;
endings[u].push_back(i);
endings[v].push_back(i);
tickets[i] = {u, v};
ticket_cost[i] = c;
}
fill(step, step + m, -1);
dfs(ROOT);
for (int i = base - 1; i > 0; --i)
t[i] = max(t[i * 2], t[i * 2 + 1]);
forn (i, m)
{
int u = tickets[i].first;
int v = tickets[i].second;
int c = ticket_cost[i];
if (is_prev(v, u))
swap(u, v);
if (is_prev(u, v))
{
assert(step[i] >= 0);
assert(is_prev(u, step[i]));
assert(is_prev(step[i], v));
u = step[i];
add(v, c);
up[u].emplace_back(v, c);
down[v].emplace_back(u, c);
}
else
{
other[u].emplace_back(v, c);
other[v].emplace_back(u, c);
}
}
go(ROOT);
cout << best << '\n';
return 0;
}








In   Python3  :





#!/bin/python3

import os
import sys
from queue import Queue
class Graph:
    def __init__(self, n, roads):
        self.n = n
        self.tree = dict()
        for i in range(n):
            self.tree[i+1] = []
        for i in range(len(roads)):
            self.tree[roads[i][0]].append((roads[i][1], roads[i][2]))
            self.tree[roads[i][1]].append((roads[i][0], roads[i][2]))

    def find_path(self, a, b, visited):
        if visited is None:
            visited = set()
        visited.add(a)
        if a==b:
            return set([a]), 0
        for c, d in self.tree[a]:
            if c not in visited:
                path, p_length = self.find_path(c, b, visited)
                if path is not None:
                    ext_path = path.copy()
                    ext_path.add(a)
                    return ext_path, p_length + d
        return None, 0



# Complete the solve function below.
def solve(roads, tickets):
    max_score = 0
    n = len(roads)+1
    g = Graph(n, roads)
    for a, b, _ in tickets:
        path, cost = g.find_path(a, b, None)
        score = - cost
        for t in tickets:
            if set(t[:2]).issubset(path):
                score += t[2]
        if score > max_score:
            max_score = score
    return max_score
            
if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    n = int(input())

    roads = []

    for _ in range(n-1):
        roads.append(list(map(int, input().rstrip().split())))

    m = int(input())

    tickets = []

    for _ in range(m):
        tickets.append(list(map(int, input().rstrip().split())))

    result = solve(roads, tickets)

    fptr.write(str(result) + '\n')

    fptr.close()








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