Submission #476028

#	Time	Username	Problem	Language	Result	Execution time	Memory
476028		CodeChamp_SS	Robot (JOI21_ho_t4)	C++17	100 / 100	1117 ms	96072 KiB

Main

#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];
map<int, vector<Edge>> cgr[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});
        cgr[u][c].pb({v, c, p}), cgr[v][c].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, cgr[cur][col]) {
                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;
}

• Subtask 1: 34 / 34
• Subtask 2: 24 / 24
• Subtask 3: 42 / 42