HackerEarth Segments maybe Super segments problem solution

In this HackerEarth Segments maybe Super segments problem solution You have an array a of integers between 0 and 10^6. Initially, you do not know exactly the value of any element of the array, but you know that the length of the array is n. Also, you are given m segments of the array l r x, where l is the left bound of the segment, r is the right bound of the segment, x is the sum of the segment. Then You are given q queries ql qr. For every query, you need to print the sum of elements of the array a on [ql,qr] interval, if it is impossible to print -1.

HackerEarth Segments maybe Super segments problem solution.

# include <bits/stdc++.h>

# include <ext/pb_ds/assoc_container.hpp>
# include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
using namespace std;
 
template<typename T> using ordered_set = tree <T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

#define _USE_MATH_DEFINES_
#define ll long long
#define ld long double
#define Accepted 0
#define pb push_back
#define mp make_pair
#define sz(x) (int)(x.size())
#define every(x) x.begin(),x.end()
#define F first
#define S second
#define lb lower_bound
#define ub upper_bound
#define For(i,x,y)  for (ll i = x; i <= y; i ++) 
#define FOr(i,x,y)  for (ll i = x; i >= y; i --)
#define SpeedForce ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0)
// ROAD to...                                                                                                                                                                                                                Red

void setIn(string s) { freopen(s.c_str(),"r",stdin); }
void setOut(string s) { freopen(s.c_str(),"w",stdout); }
void setIO(string s = "") {
    // cin.exceptions(cin.failbit); 
    // throws exception when do smth illegal
    // ex. try to read letter into int
    if (sz(s)) { setIn(s+".in"), setOut(s+".out"); }
}

const double eps = 0.000001;
const ld pi = acos(-1);
const int maxn = 1e7 + 9;
const int mod = 1e9 + 7;
const ll MOD = 1e18 + 9;
const ll INF = 1e18 + 123;
const int inf = 2e9 + 11;
const int mxn = 1e6 + 9; 
const int N = 2e5+5;                                         
const int M = 22;
const int pri = 997;
const int Magic = 2101;

const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, -1, 0, 1};
mt19937 gen(chrono::steady_clock::now().time_since_epoch().count());
 
int rnd (int l, int r) {
    return uniform_int_distribution<int> (l, r)(gen);
}
int n, m, q;
vector < pair<int, ll> > g[N];
ll f[N];
int col[N];

void dfs (int v, int cc) {
    col[v] = cc;
    
    for (auto e : g[v]) {
        int to = e.first;
        ll sum = e.second;
        if(to < v) sum *= -1;
        
        if(col[to] != -1) {
            assert(f[to] - f[v] == sum);
            continue;
        }
        
        f[to] = f[v] + sum;
        dfs(to, cc);
    }
}
    
void solve() {
    cin >> n >> m >> q;
    assert(1 <= n && n <= 200000);
    assert(1 <= m && m <= 200000);
    assert(1 <= q && q <= 200000);
    for (int i = 0; i <= n; ++i) {
        g[i].clear();
        col[i] = -1;
    }
    
    for (int i = 1; i <= m; ++i) {
        int l, r, x;
        cin >> l >> r >> x;
        --l;
        g[l].pb({r, x});
        g[r].pb({l, x});
    }
    int cur = 0;
    for (int i = 0; i <= n; ++i) if(col[i] == -1) {
        f[i] = 0;
        dfs(i, ++cur);
    }
    
    for (int i = 1; i <= q; ++i) {
        int ql, qr;
        cin >> ql >> qr;
        --ql;
        if(col[ql] != col[qr]) {
            cout << "-1n";
            continue;
        }
        
        cout << f[qr] - f[ql] << 'n';
    }
}

int main () {
    SpeedForce;
    //setIO("test1");
    int T = 1;
    cin >> T;
    while(T--) solve();
    
    return Accepted;
}

HackerEarth Segments maybe Super segments problem solution.

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