Submission #534840

#TimeUsernameProblemLanguageResultExecution timeMemory
534840Haruto810198Olympic Bus (JOI20_ho_t4)C++17
0 / 100
34 ms1868 KiB
#include <bits/stdc++.h> using namespace std; #define int long long #define double long double #define FOR(i, l, r, d) for(int i=(l); i<=(r); i+=(d)) #define szof(x) ((int)(x).size()) #define vi vector<int> #define pii pair<int, int> #define F first #define S second #define pb push_back #define eb emplace_back #define mkp make_pair const int INF = INT_MAX; const int LNF = 1e18; const int MOD = 1000000007; const int mod = 998244353; const double eps = 1e-12; //#pragma GCC optimize("Ofast") //#pragma GCC optimize("unroll-loops") const int MAX = 210; int n, m; int from[MAX], to[MAX], c[MAX], d[MAX]; // information of edges vi edge[MAX]; vi spe; // edges on shortest path s -> t bool isspe[MAX]; int dis[MAX][MAX], dis2[MAX][MAX]; // dis2 : dis. without using edges in sp int res[MAX]; // res[i] = optimal answer if we choose to reverse edge i void Dijkstra(int s, int t){ int disDij[MAX]; bool vis[MAX]; int pv[MAX], pe[MAX]; // id. of previous vertex, edge FOR(i, 1, n, 1){ disDij[i] = LNF + 1; vis[i] = 0; } disDij[s] = 0; FOR(owo, 1, n, 1){ int Min = LNF, u = 0; FOR(i, 1, n, 1){ if(vis[i] == 0 and Min > disDij[i]){ Min = disDij[i]; u = i; } } for(int e : edge[u]){ int v = to[e]; if(disDij[u] + c[e] < disDij[v]){ disDij[v] = disDij[u] + c[e]; pv[v] = u; pe[v] = e; } } vis[u] = 1; /* cerr<<"dis : "; FOR(i, 1, n, 1){ if(disDij[i] > 100) cerr<<"- "; else cerr<<disDij[i]<<" "; } cerr<<endl; */ } // find shortest path s -> t spe.clear(); FOR(i, 1, m, 1){ isspe[i] = 0; } int ptr = t; while(ptr != s){ spe.pb(pe[ptr]); isspe[pe[ptr]] = 1; ptr = pv[ptr]; } reverse(spe.begin(), spe.end()); assert(!spe.empty()); } void FloydWarshall(){ // init. dis, dis2 FOR(i, 1, n, 1){ FOR(j, 1, n, 1){ dis[i][j] = dis2[i][j] = LNF; } dis[i][i] = dis2[i][i] = 0; } FOR(e, 1, m, 1){ int u = from[e], v = to[e]; dis[u][v] = min(dis[u][v], c[e]); if(!isspe[e]) dis2[u][v] = min(dis2[u][v], c[e]); } FOR(k, 1, n, 1){ FOR(i, 1, n, 1){ FOR(j, 1, n, 1){ dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]); dis2[i][j] = min(dis2[i][j], dis2[i][k] + dis2[k][j]); } } } } void solve(int s, int t){ // (s, t) = (1, n) or (n, 1) //cerr<<"solve "<<s<<" -> "<<t<<endl; Dijkstra(s, t); //cerr<<"Dij : ok"<<endl; FloydWarshall(); //cerr<<"FW : ok"<<endl; // e not on sp FOR(e, 1, m, 1){ if(isspe[e]) continue; int u = from[e], v = to[e]; res[e] += min(dis[s][t], dis[s][v] + c[e] + dis[u][t]); } // e on sp int sz = szof(spe); // FOR(i, 0, sz-2, 1){ assert(to[spe[i]] == from[spe[i+1]]); } // FOR(i, 0, sz-1, 1){ int e = spe[i]; int Min = LNF; FOR(j, 0, i, 1){ FOR(k, i, sz-1, 1){ int u = from[spe[j]], v = to[spe[k]]; Min = min(Min, dis[s][u] + dis2[u][v] + dis[v][t]); } } res[e] += Min; } } signed main(){ ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0); cin>>n>>m; FOR(i, 1, m, 1){ cin>>from[i]>>to[i]>>c[i]>>d[i]; edge[from[i]].pb(i); } //cerr<<"in : ok"<<endl; // build 2 extra edges to guarantee there is a path 1 -> n, n -> 1 from[m+1] = 1, from[m+2] = n; to[m+1] = n, to[m+2] = 1; c[m+1] = c[m+2] = LNF; d[m+1] = d[m+2] = LNF; edge[1].pb(m+1); edge[n].pb(m+2); m += 2; solve(1, n); solve(n, 1); int RES = dis[1][n] + dis[n][1]; // don't flip edges FOR(e, 1, m, 1){ res[e] += d[e]; RES = min(RES, res[e]); // flip e } if(RES < LNF) cout<<RES<<'\n'; else cout<<-1<<'\n'; // subtask 2 FOR(i, 1, m-2, 2){ assert(res[i] == res[i+1]); } return 0; }
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