제출 #1130437

#제출 시각아이디문제언어결과실행 시간메모리
1130437enzyCommuter Pass (JOI18_commuter_pass)C++20
0 / 100
36 ms2368 KiB
#include <bits/stdc++.h> #define int long long using namespace std; const int maxn=310; const int inf=1e18+10; vector<pair<int,int>>adj[maxn]; int dist[maxn][4], n, marc[maxn], best, dp[maxn][maxn]; void dijkstra(int x, int t,int dir){ for(int i=1;i<=n;i++) dist[i][t]=inf; dist[x][t]=0; set<pair<int,int>>s; for(int i=1;i<=n;i++) s.insert({dist[i][t],i}); while(!s.empty()){ auto f=s.begin(); int u=f->second; s.erase(f); for(auto p : adj[u]){ int viz=p.first, w=p.second; bool flag=false; if(dist[u][2]+w+dist[viz][3]==best||dist[u][3]+w+dist[viz][2]==best) flag=true; if(flag&&dist[u][dir]+w==dist[viz][dir]&&dist[u][t]<dist[viz][t]){ // estou usando uma aresta já DIRECIONADA do commuter pass, então passo de graça s.erase({dist[viz][t],viz}); dist[viz][t]=dist[u][t]; s.insert({dist[viz][t],viz}); } if(dist[u][t]+w<dist[viz][t]){ // estou utilizando uma aresta normalmente, então tenho que pagar s.erase({dist[viz][t],viz}); dist[viz][t]=dist[u][t]+w; s.insert({dist[viz][t],viz}); } } } } signed main() { ios_base::sync_with_stdio(false); cin.tie(NULL); int m, s, t, u, v; cin >> n >> m >> s >> t >> u >> v; assert(n<=300); for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) dp[i][j]=inf; for(int i=1;i<=m;i++){ int a, b, c; cin >> a >> b >> c; adj[a].push_back({b,c}); adj[b].push_back({a,c}); dp[a][b]=c; dp[b][a]=c; } for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) for(int k=1;k<=n;k++) dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]); int ans=dp[u][v]; for(int i=1;i<=n;i++) for(int j=1;j<=n;j++){ if(dp[s][i]+dp[i][j]+dp[j][t]==dp[s][t]) ans=min({ans,dp[u][i]+dp[j][v],dp[v][i]+dp[j][u]}); } dijkstra(s,2,0); dijkstra(t,3,0); best=dist[t][2]; for(int i=1;i<=n;i++) if(dist[i][2]+dist[i][3]==best) marc[i]++; dijkstra(u,0,2); // direcionando todas as arestas do commuter pass de S->T dijkstra(u,1,3); // direcionando todas as arestas do commuter pass de T->S int resp=min(dist[v][0],dist[v][1]); cout << ans << endl; //assert(resp>=ans); //cout << resp << endl; return 0; }// sub3
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