Submission #777444

#TimeUsernameProblemLanguageResultExecution timeMemory
777444OrazBTracks in the Snow (BOI13_tracks)C++17
82.50 / 100
2095 ms43424 KiB
#include <bits/stdc++.h> #include <ext/pb_ds/assoc_container.hpp> #include <ext/pb_ds/tree_policy.hpp> #include <functional> using namespace __gnu_pbds; using namespace std; typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set; //Dijkstra->set //set.find_by_order(x) x-position value //set.order_of_key(x) number of strictly less elements don't need *set.?? #define N 100005 #define wr cout << "Continue debugging\n"; #define all(x) (x).begin(), (x).end() #define ll long long int #define pii pair <int, int> #define pb push_back #define ff first #define ss second int X[] = {-1, 1, 0, 0}; int Y[] = {0, 0, -1, 1}; int main () { ios::sync_with_stdio(false); cin.tie(0); int n, m; cin >> n >> m; vector<vector<char>> c(n+1, vector<char>(m+1)); for (int i = 1; i <= n; i++){ for (int j = 1; j <= m; j++) cin >> c[i][j]; } int ans = 0; vector<vector<bool>> vis(n+1, vector<bool>(m+1, 0)); priority_queue<pair<int,pii>> q; q.push({-1, {1, 1}}); while(!q.empty()){ int x = q.top().ss.ff, y = q.top().ss.ss, w = q.top().ff; q.pop(); if (vis[x][y]) continue; vis[x][y] = 1; ans = max(ans, -w); for (int i = 0; i < 4; i++){ int a = x+X[i], b = y+Y[i]; if (a > 0 and b > 0 and a <= n and b <= m and !vis[a][b] and c[a][b] != '.'){ if (c[x][y] != c[a][b]) q.push({w-1, {a, b}}); else q.push({w, {a, b}}); } } } cout << ans << '\n'; }
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