**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 :

` ````
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()
```

## View More Similar Problems

## Mr. X and His Shots

A cricket match is going to be held. The field is represented by a 1D plane. A cricketer, Mr. X has N favorite shots. Each shot has a particular range. The range of the ith shot is from Ai to Bi. That means his favorite shot can be anywhere in this range. Each player on the opposite team can field only in a particular range. Player i can field from Ci to Di. You are given the N favorite shots of M

View Solution →## Jim and the Skyscrapers

Jim has invented a new flying object called HZ42. HZ42 is like a broom and can only fly horizontally, independent of the environment. One day, Jim started his flight from Dubai's highest skyscraper, traveled some distance and landed on another skyscraper of same height! So much fun! But unfortunately, new skyscrapers have been built recently. Let us describe the problem in one dimensional space

View Solution →## Palindromic Subsets

Consider a lowercase English alphabetic letter character denoted by c. A shift operation on some c turns it into the next letter in the alphabet. For example, and ,shift(a) = b , shift(e) = f, shift(z) = a . Given a zero-indexed string, s, of n lowercase letters, perform q queries on s where each query takes one of the following two forms: 1 i j t: All letters in the inclusive range from i t

View Solution →## Counting On a Tree

Taylor loves trees, and this new challenge has him stumped! Consider a tree, t, consisting of n nodes. Each node is numbered from 1 to n, and each node i has an integer, ci, attached to it. A query on tree t takes the form w x y z. To process a query, you must print the count of ordered pairs of integers ( i , j ) such that the following four conditions are all satisfied: the path from n

View Solution →## Polynomial Division

Consider a sequence, c0, c1, . . . , cn-1 , and a polynomial of degree 1 defined as Q(x ) = a * x + b. You must perform q queries on the sequence, where each query is one of the following two types: 1 i x: Replace ci with x. 2 l r: Consider the polynomial and determine whether is divisible by over the field , where . In other words, check if there exists a polynomial with integer coefficie

View Solution →## Costly Intervals

Given an array, your goal is to find, for each element, the largest subarray containing it whose cost is at least k. Specifically, let A = [A1, A2, . . . , An ] be an array of length n, and let be the subarray from index l to index r. Also, Let MAX( l, r ) be the largest number in Al. . . r. Let MIN( l, r ) be the smallest number in Al . . .r . Let OR( l , r ) be the bitwise OR of the

View Solution →