제출 #757665

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
757665		sysia	Olympic Bus (JOI20_ho_t4)	C++17	37 / 100	1062 ms	3784 KiB

ho_t4

//Sylwia Sapkowska
#include <bits/stdc++.h>
#pragma GCC optimize("O3", "unroll-loops")
using namespace std;
 
void __print(int x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << "'" << x << "'";}
void __print(const char *x) {cerr << '"' << x << '"';}
void __print(const string &x) {cerr << '"' << x << '"';}
void __print(bool x) {cerr << (x ? "true" : "false");}
 
template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ", "; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? ", " : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifdef LOCAL
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif
#define ll long long
typedef pair<int, int> T;
set<T>cycle;
const int oo2 = 2e9+7;
const int mx = 202;
int n;
int pre[mx];
priority_queue<T>pq;
 
vector<int> dijkstra(vector<int>&dist, vector<vector<T>>&g, int from, int to = -1){
    dist[from] = 0;
    
    if (to != -1) {
        for (int i = 1; i<=n; i++) pre[i] = 0;
    }
    pq.push({0, from});
    while ((int)pq.size()){
        auto [d, v] = pq.top();
        pq.pop();
        if (dist[v] < d) continue;
        for (auto [x, c]: g[v]){
            if (dist[x] > dist[v] + c){
                pre[x] = v;
                dist[x] = dist[v] + c;
                pq.push({dist[x], x});
            }
        }
    }
    if (to != -1){
        int now = to;
        int prev = -1;
        while (now){
            if (prev != -1) cycle.insert({now, prev});
            prev = now;
            now = pre[now];
        }
    }
    return dist;
}


int dist[mx];
int dijkstra2(vector<vector<T>>&g, int from, int to){
    for (int i = 1; i<=n; i++) dist[i] = oo2;
    dist[from] = 0;
    pq.push({0, from});
    while ((int)pq.size()){
        auto [d, v] = pq.top(); pq.pop();
        if (dist[v] < d) continue;
        for (auto [x, c]: g[v]){
            if (dist[x] > dist[v] + c){
                dist[x] = dist[v] + c;
                pq.push({dist[x], x});
            }
        }
    }
    return dist[to];
}
 

void solve(){
    int m; cin >> n >> m;
    vector<vector<T>>g(n+1), gt(n+1);
    vector<tuple<int, int, int, int>>edges;
    for (int i = 0; i<m; i++){
        int a, b, c, d; cin >> a >> b >> c >> d;
        g[a].emplace_back(b, c);
        gt[b].emplace_back(a, c);
        edges.emplace_back(a, b, c, d);
    }
    vector<int>d1(n+1, oo2), dn(n+1, oo2), d1rev(n+1, oo2), dnrev(n+1, oo2);
    dijkstra(d1, g, 1, n); //1 do x
    dijkstra(dn, g, n, 1); //n do x
    dijkstra(d1rev, gt, 1); //x do 1
    dijkstra(dnrev, gt, n); //x do n
    long long ans = (long long)d1[n] + dn[1];
    int k = 0;
    debug(cycle);
    for (auto &[a, b, c, d]: edges){
        if (cycle.find({a, b}) != cycle.end()){
            debug(a, b);
            auto [A, B, C, D] = edges[k];
            for (int i = 0; i<(int)g[A].size(); i++){
                if (g[A][i] == make_pair(B, C)){
                    swap(g[A][i], g[A].back());
                    break;
                }
            }
            g[A].pop_back();
            g[B].emplace_back(A, C);
            long long tmp = (long long)d + (long long)dijkstra2(g, 1, n) + (long long)dijkstra2(g, n, 1);
            debug(tmp);
            g[B].pop_back();
            g[A].emplace_back(B, C);
            ans = min(ans, tmp);
        } else {
            ans = min(ans, (ll)d + min((ll)d1[n], (ll)d1[b]+c+dnrev[a]) + min((ll)dn[1], (ll)dn[b] + c + d1rev[a]));
        }
        k++;
    }
    if (ans >= oo2) cout << "-1\n";
    else cout << ans << "\n";
}
 
int32_t main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
 
    solve();
 
    return 0;
}

• Subtask 1: 5 / 5
• Subtask 2: 11 / 11
• Subtask 3: 21 / 21
• Subtask 4: 0 / 63