This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#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;
#define ff first
#define ss second
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define setbits(x) __builtin_popcountll(x)
#define zrobits(x) __builtin_ctzll(x)
#define sz(v) (int)v.size()
#define ps(y) cout << fixed << setprecision(y)
#define ms(arr, v) memset(arr, v, sizeof(arr))
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
#define trav(x, v) for(auto &x: v)
#define w(t) int t; cin >> t; while(t--)
#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 rep0(i, n) rep(i, 0, n - 1)
#define rrep0(i, n) rrep(i, n - 1, 0)
#define rep1(i, n) rep(i, 1, n)
#define rrep1(i, n) rrep(i, n, 1)
#define inp(arr, n) rep0(i, n) cin >> arr[i];
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<ll, ll> pii;
typedef vector<ll> vi;
typedef vector<vi> vvi;
typedef vector<pii> vp;
typedef vector<bool> vb;
typedef vector<string> vs;
typedef map<ll, ll> mii;
typedef map<char, ll> mci;
typedef priority_queue<ll> pq_mx;
typedef priority_queue<tuple<ll, int, int>, vector<tuple<ll, int, int>>, greater<>> pq_mn;
typedef tree<ll, null_type, less<>, rb_tree_tag, tree_order_statistics_node_update> pbds;
/*
* find_by_order(i) -> returns an iterator to the element at ith position (0 based)
* order_of_key(i) -> returns the position of element i (0 based)
*/
const int N = 1e5 + 5;
const int mod = 1e9 + 7;
//const int mod = 998244353;
const ll inf = 1e18;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void fio() {
ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
}
struct Edge {
int v, c;
ll p;
};
int n, m;
ll dp[N];
vector<Edge> gr[N];
mii dp1[N], pre[N];
int main() {
fio();
cin >> n >> m;
rep0(x, m) {
int u, v, c;
ll p;
cin >> u >> v >> c >> p;
gr[u].pb({v, c, p}), gr[v].pb({u, c, p});
pre[u][c] += p, pre[v][c] += p;
}
rep0(x, n + 1) dp[x] = inf;
pq_mn q;
q.push({0, 1, 0}), dp[1] = 0;
while (!q.empty()) {
auto[cost, cur, col] = q.top();
q.pop();
if (col) {
if (dp1[cur][col] < cost) continue;
trav(neigh, gr[cur]) {
if (neigh.c != col) continue;
// we can't flip the same edge again
ll curCost = pre[cur][neigh.c] - neigh.p + cost;
if (curCost < dp[neigh.v]) {
dp[neigh.v] = curCost;
q.push({dp[neigh.v], neigh.v, 0});
}
}
} else {
if (dp[cur] < cost) continue;
trav(neigh, gr[cur]) {
// we don't flip this edge
ll curCost = pre[cur][neigh.c] - neigh.p + cost;
if (curCost < dp[neigh.v]) {
dp[neigh.v] = curCost;
q.push({dp[neigh.v], neigh.v, 0});
}
// we flip this edge but not others with the same color
curCost = neigh.p + cost;
if (curCost < dp[neigh.v]) {
dp[neigh.v] = curCost;
q.push({dp[neigh.v], neigh.v, 0});
}
// we don't include this edge and robot will leave 'cur' with some other edge having color 'col'
curCost = cost;
if (!dp1[neigh.v].count(neigh.c) or dp1[neigh.v][neigh.c] > curCost) {
dp1[neigh.v][neigh.c] = curCost;
q.push({curCost, neigh.v, neigh.c});
}
}
}
}
cout << (dp[n] == inf ? -1 : dp[n]);
return 0;
}
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