제출 #511681

#제출 시각아이디문제언어결과실행 시간메모리
511681Tizariox악어의 지하 도시 (IOI11_crocodile)C++14
0 / 100
5 ms5196 KiB
#include <bits/stdc++.h> using namespace std; using ll = long long; using pii = pair<int, int>; #define FOR(i, a, b) for (int i = a; i < (b); i++) #define F0R(i, a) for (int i = 0; i < (a); i++) #define F0R1(i, a) for (int i = 1; i <= (a); i++) #define FORd(i, a, b) for (int i = (b)-1; i >= a; i--) #define F0Rd(i, a) for (int i = (a)-1; i >= 0; i--) #define F0Rd1(i, a) for (int i = a; i > 0; i--) #define SORT(vec) sort(vec.begin(), vec.end()) #define S0RT(arr, n) sort(arr, arr + n) #define pA first #define pB second #define MOD 1000000007 #define nl "\n" #define pb push_back int n, m, k; vector<int> exits; int pathInd = 0; // index each path to make sure we don't duplicate paths class cmp { public: bool operator()(vector<int> a, vector<int> b) { if (a[1] == b[1]) { return a[2] > b[2]; } return a[1] > b[1]; } }; pair<pii, pii> dist[100000]; // 2 smallest distances to each node vector<pii> adj[100000]; bool shortestVis[100000]; // whether the shortest path for each node has already been taken void dijkstra() { // traceback for (int i = 0; i < n; i++) { dist[i] = {{1e9, -1}, {1e9, -1}}; shortestVis[i] = false; } priority_queue<vector<int>, vector<vector<int>>, cmp> nodes; for (int root : exits) { dist[root] = {{0, pathInd}, {0, pathInd}}; nodes.push({root, 0, pathInd++}); shortestVis[root] = true; } while (!nodes.empty()) { int node = nodes.top()[0]; int minDist = nodes.top()[1]; int pathInd = nodes.top()[2]; nodes.pop(); if (!shortestVis[node]) { shortestVis[node] = true; if (pathInd != dist[node].pA.pB) { dist[node].pA = {minDist, pathInd}; } continue; } else if (minDist > dist[node].pB.pA) { continue; } else if (pathInd != dist[node].pB.pB) { assert(minDist > dist[node].pB.pA); // assert((minDist > dist[node].pB.pA) || (pathInd > dist[node].pB.pB)); continue; } for (pii i : adj[node]) { int neighbor = i.pA; int length = i.pB; if (minDist + length < dist[neighbor].pA.pA) { dist[neighbor].pB = dist[neighbor].pA; dist[neighbor].pA = {minDist + length, pathInd}; nodes.push({neighbor, dist[neighbor].pA.pA, pathInd++}); } else if (minDist + length < dist[neighbor].pB.pA) { dist[neighbor].pB = {minDist + length, pathInd}; nodes.push({neighbor, dist[neighbor].pB.pA, pathInd++}); } } } } int travel_plan(int n1, int m1, int (*r1) [2], int* L, int k1, int* P) { n = n1; m = m1; k = k1; for (int i = 0; i < m; i++) { int a, b, c; a = r1[i][0]; b = r1[i][1]; c = L[i]; adj[a].pb(pii(b, c)); adj[b].pb(pii(a, c)); } FOR(i, 0, k) { exits.pb(P[i]); } dijkstra(); return dist[0].pB.pA; } // int main() { // ios_base::sync_with_stdio(0); // cin.tie(0); // cin >> n >> m >> k; // for (int i = 0; i < m; i++) { // int a, b, c; // cin >> a >> b >> c; // adj[a].pb(pii(b, c)); // adj[b].pb(pii(a, c)); // } // FOR(i, 0, k) { // int num; // cin >> num; // exits.pb(num); // } // dijkstra(); // cout << dist[0].pB.pA << nl; // } // 5 7 2 // 0 2 4 // 0 3 3 // 3 2 2 // 2 1 10 // 0 1 100 // 0 4 7 // 3 4 9 // 1 3
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