제출 #305359

#제출 시각아이디문제언어결과실행 시간메모리
305359ChrisGe123Commuter Pass (JOI18_commuter_pass)C++14
0 / 100
2100 ms51220 KiB
#include <bits/stdc++.h> using namespace std; typedef long long ll; #define maxn 100002 #define maxm 200002 int n, m, s, t, u, v; vector<int> parents[maxn]; //the first one stores the parents using the dijkstras thing for source S, the second one stores with source T vector<pair<int, ll> > adj[maxn]; //first one is the id, second one is the cost of the edge //max cost is one billion ll dist[maxn][3]; //the first one stores the distances from source S, the second stores from Source U, the third one stores from Source V vector<int> shortpathchildren[maxn]; vector<int> dagnodes; bool vis[maxn]; ll dpU[maxn][2]; //dpU[i] gives the minimum distance from U to any vertex between i and S ll dpV[maxn][2]; //dpV[i] gives the minimum distance from V to any vertex between i and T ll dpUfuncone(int i) { if (dpU[i][0] != LLONG_MAX) { return dpU[i][0]; } if (i == s) { return dist[s][1]; } ll temp = dist[i][1]; for (auto j: parents[i]) { temp = min(temp, dpUfuncone(j)); } // if (temp < 0) // { // cerr << i << " " << temp << endl; // } return dpU[i][0] = temp; } ll dpVfuncone(int i) { if (dpV[i][0] != LLONG_MAX) { return dpV[i][0]; } if (i == t) { //cerr << "YES: " << dist[t][2] << endl; return dist[t][2]; } ll temp = dist[i][2]; for (auto j: shortpathchildren[i]) { temp = min(temp, dpVfuncone(j)); } return dpV[i][0] = temp; } ll dpUfunctwo(int i) { if (dpU[i][1] != LLONG_MAX) { return dpU[i][1]; } if (i == t) { return dist[t][1]; } ll temp = dist[i][1]; for (auto j: shortpathchildren[i]) { temp = min(temp, dpUfunctwo(j)); } return dpV[i][1] = temp; } ll dpVfunctwo(int i) { if (dpU[i][1] != LLONG_MAX) { return dpU[i][0]; } if (i == s) { return dist[s][2]; } ll temp = dist[i][2]; for (auto j: parents[i]) { temp = min(temp, dpVfunctwo(j)); } return dpV[i][1] = temp; } void dijkstras(int source, int use) { priority_queue<pair<int, ll>, vector<pair<int, ll> >, greater<pair<int, ll> > > points; //first one is the id, second is the distance from source for (int i=0; i<n; i++) { dist[i][use] = LLONG_MAX; } dist[source][use] = 0; points.push(make_pair(source, 0)); while (points.size()) { pair<int, ll> temp = points.top(); points.pop(); ll cdist = temp.second; int id = temp.first; // if (use == 2) // { // cerr << id << " " << cdist << endl; // // if (id == 2) // // { // // cerr << cdist << endl; //for some reason this becomes -1294967296 // // cerr << dist[id][use] << endl; // // } // } if (cdist != dist[id][use]) { continue; } for (auto k: adj[id]) { // if (use == 2) // { // cerr << id << ": " << k.first << endl; // // cerr << dist[k.first][use] << endl; // // cerr << cdist + k.second << endl; // // cerr << endl; // } if (dist[k.first][use] > cdist + k.second) { // if (use == 2) // { // cerr << id << ": " << k.first << endl; // // cerr << dist[k.first][use] << endl; // // cerr << cdist + k.second << endl; // // cerr << endl; // } dist[k.first][use] = cdist + k.second; if (use == 0) { parents[k.first].clear(); parents[k.first].push_back(id); } // if (use == 2 && k.first == 2) // { // cerr << dist[k.first][use] << endl; //this is still 3000000000 // } points.push(make_pair(k.first, dist[k.first][use])); } else if (use == 0 && dist[k.first][use] == cdist + k.second) { parents[k.first].push_back(id); //we won't push this back into the queue because if the thing is not infinity, that the last time that it was reduced // it was already pushed into the priority queue, meaning that another update with still the same distance is meaningless } } } } int main() { ios_base::sync_with_stdio(false); cin.tie(0); //freopen("joi.in", "r", stdin); //freopen("joi.out", "w", stdout); cin >> n >> m >> s >> t >> u >> v; s--; t--; u--; v--; for (int i=0; i<m; i++) { int a, b, c; cin >> a >> b >> c; a--; b--; adj[a].push_back(make_pair(b,c)); adj[b].push_back(make_pair(a,c)); } //let's first see that we've read everything in correctly dijkstras(s, 0); //then check that we're doing the dijkstras correctly from source s dijkstras(u, 1); dijkstras(v, 2); //cerr << dist[t][2] << endl; //then check that we're getting the right dijkstras from u and v //we should probably assemble the lowest path tree queue<int> stuff; stuff.push(t); vis[t] = true; dagnodes.push_back(t); while (stuff.size()) { int cur = stuff.front(); stuff.pop(); for (int i=0; i<parents[cur].size(); i++) { int k = parents[cur][i]; shortpathchildren[k].push_back(cur); if (!vis[k]) { dagnodes.push_back(k); stuff.push(k); vis[k] = true; } } } //then we should check that the dagnodes and children are correct for (auto k: dagnodes) { dpU[k][0] = LLONG_MAX; dpV[k][0] = LLONG_MAX; } ll minlengthone = LLONG_MAX; for (auto k: dagnodes) { // ll x = dpUfuncone(k); // ll y = dpVfuncone(k); // cerr << "from node " << k << ", we get " << x << " " << y << endl; minlengthone = min(minlengthone, dpUfuncone(k) + dpVfuncone(k)); // if (dpUfuncone(k) + dpVfuncone(k) < 0) // { // cerr << k << endl; // } } //then we check the dp process for (auto k: dagnodes) { dpU[k][1] = LLONG_MAX; dpV[k][1] = LLONG_MAX; } ll minlengthtwo = LLONG_MAX; for (auto k: dagnodes) { minlengthtwo = min(minlengthtwo, dpUfunctwo(k) + dpVfunctwo(k)); } //then we check the second dp process // cerr << dist[u][2] << endl; // cerr << minlengthone << endl; // cerr << minlengthtwo << endl; // cerr << LLONG_MIN << endl; //how did both become LL_ONG_MIN? this must mean that dpUfunc and dpVfunc returned LLONG_MIN, but what? cout << min(dist[u][2], min(minlengthone, minlengthtwo)) << endl; }

컴파일 시 표준 에러 (stderr) 메시지

commuter_pass.cpp: In function 'int main()':
commuter_pass.cpp:198:18: warning: comparison of integer expressions of different signedness: 'int' and 'std::vector<int>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
  198 |   for (int i=0; i<parents[cur].size(); i++)
      |                 ~^~~~~~~~~~~~~~~~~~~~
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