제출 #926787

#제출 시각아이디문제언어결과실행 시간메모리
926787lumid사이버랜드 (APIO23_cyberland)C++17
0 / 100
3131 ms1325012 KiB
// Source: https://usaco.guide/general/io #include <algorithm> #include <bits/stdc++.h> #include <climits> #include <iterator> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef vector<int> vi; typedef pair<int, int> ii; typedef vector<ii> vii; typedef long double ld; typedef pair<ll, ll> pll; #define FOR(i, a, b) for(int i = a; i < b; i++) #define ROF(i, a, b) for(int i = a; i >= b; i--) #define ms memset #define pb push_back #define fi first #define se second #define inp(n, a) vector<ll> a;for(int i=0;i<n;i++){ll now;cin>>now;a.pb(now);} #define all(a) a.begin(),a.end() #define show(a) for(long long loppls=0;loppls<(long long)(a.size()-1);loppls++)cout<<a[loppls]<<' ';cout<<a[a.size()-1]; #ifdef reimufumo #define owo(x) std::cerr << x; #define ovo(a) for(long long loppls=0;loppls<(long long)(a.size()-1);loppls++)cerr<<a[loppls]<<' ';cerr<<a[a.size()-1]; #define ouo(a,x) for(long long loppls=0;loppls<x-1;loppls++)cerr<<a[loppls];cerr<<a[x-1]; #define dbg(x) std::cerr << #x << " = " << (x) << " (L" << __LINE__ << ") " << __FILE__ << std::endl #define dbgif(cond, x) ((cond) ? std::cerr << #x << " = " << (x) << " (L" << __LINE__ << ") " << __FILE__ << std::endl : std::cerr) #else #define owo(x) ((void)0) #define ovo(a) ((void)0) #define ouo(a,x) ((void)0) #define dbg(x) ((void)0) #define dbgif(cond, x) ((void)0) #endif long long binpow(long long a, long long b, long long m) { a %= m; long long res = 1; while (b > 0) { if (b & 1) res = res * a % m; a = a * a % m; b >>= 1; } return res; } long long inv(long long a, long long p){ return binpow(a, p-2, p); } vector<ll> fact; // must be init if nCk needed long long nCk(long long n, long long k, long long p){ return ((fact[n] * inv(fact[k], p) % p) * inv(fact[n-k], p)) % p; } ll sum2(ll a, ll l, ll n){return (n*(a+l))/2;} ll ceil2(ll a, ll b){ll c=a/b;if(a%b!=0)c++;return c;} ll floor2(ll x, ll m){ll r=(x%m+m)%m;return (x-r)/m;} const ll INF=1e16,MAX=100020,MOD=998244353; ll n,m,k,h; vector<vector<vector<ll>>> adj; // adj[u] {v,w,type} vector<double> dist; vector<bool> processed; void init(){ for(int i=0;i<=(k+1)*n;i++){ adj.pb({}); dist.pb(INF); // check on this because of the double processed.pb(0); } } double calc(double d,ll w,ll typ){ // todo: calculate distance if(typ==1){ return (double)(d)+(double)(w); }else if(typ==0){ return 0.0; }else{ return (double)(((double)(d)+(double)(w))/2.0); } } void dijkstra(){ priority_queue<pair<double,ll>> q; q.push({0,0}); dist[0]=0; while(!q.empty()){ ll u=q.top().se;q.pop(); if(processed[u])continue; processed[u]=1; for(vector<ll> node:adj[u]){ ll v=node[0],w=node[1],typ=node[2]; double distnow=calc(dist[u],w,typ); if(distnow<dist[v]){ dist[v]=distnow; q.push({-dist[v],v}); } } } } double solve(int N, int M, int K, int H, std::vector<int> x, std::vector<int> y, std::vector<int> c, std::vector<int> arr){ // THIS PROBLEM IS 0-INDEXED n=N,m=M,k=min(200,K),h=H; init(); for(int i=0;i<m;i++){ // whaaaaaaaaat ll u=x[i],v=y[i],w=c[i]; ll typ1=arr[u],typ2=arr[v]; for(int j=0;j<k;j++){ if(typ1==0) adj[j*n+v].pb({j*n+u,0,0}); else adj[j*n+v].pb({j*n+u,w,1}); if(typ2==0) adj[j*n+u].pb({j*n+v,0,0}); else adj[j*n+u].pb({j*n+v,w,1}); if(typ1==2) adj[j*n+v].pb({(j+1)*n+u,w,2}); if(typ2==2) adj[j*n+u].pb({(j+1)*n+v,w,2}); } if(typ1==0) adj[k*n+v].pb({k*n+u,0,0}); else adj[k*n+v].pb({k*n+u,w,1}); if(typ2==0) adj[k*n+u].pb({k*n+v,0,0}); else adj[k*n+u].pb({k*n+v,w,1}); } // rmb to join down fencengtu for(int i=0;i<k;i++){ for(int j=0;j<n;j++){ adj[i*n+j].pb({(i+1)*n+j,0,1}); } } dijkstra(); return (dist[k*n+h]>=1e12?-1:dist[k*n+h]); // beware of double precision } #ifdef reimufumo int main() { ios::sync_with_stdio(false); cin.tie(NULL); int ninp,minp,kinp,hinp;cin>>ninp>>minp>>kinp>>hinp; vector<int> xinp,yinp,cinp,arrinp; for(int i=0;i<ninp;i++){ int arrnow;cin>>arrnow; arrinp.pb(arrnow); } for(int i=0;i<minp;i++){ int xnow,ynow,cnow;cin>>xnow>>ynow>>cnow; xinp.pb(xnow);yinp.pb(ynow);cinp.pb(cnow); } cout<<solve(ninp,minp,kinp,hinp,xinp,yinp,cinp,arrinp); for(int i=0;i<=k;i++){ for(int j=0;j<n;j++){ cerr<<dist[i*n+j]<<' '; }cerr<<'\n'; } } #endif
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