제출 #1062365

#제출 시각아이디문제언어결과실행 시간메모리
1062365danielzhuTracks in the Snow (BOI13_tracks)C++17
100 / 100
437 ms28312 KiB
#include <bits/stdc++.h> using namespace std; #define MX 4001 typedef long long LL; typedef pair<int, int> PR; typedef tuple<int, int, int> TP; const int MOD = (int)1e9 + 7; const int INF = (int)1e9 + 5; const LL MAXV = (LL)1e18; using vi = vector<int>; using vii = vector<vi>; int H, W; string md[MX]; //snow meadow //idea: the footprints (not all) of an animal can be covered by all its successors' footprints. //Also can be solved using Disjoint Sets instead of BFS/DFS? int solve() { cin >> H >> W; //read the snow meadow info for (int i = 0; i < H; i++) cin >> md[i]; queue<PR> Q[2]; //two queues for BFS responding to R and F int ans = 0; if (md[0][0] == '.') return ans; vector<char> animal(1, md[0][0]); animal.push_back(md[0][0] == 'F' ? 'R' : 'F'); //alternate animal's footprints to discover int cur = 0; //an index of animial to reveal the current animal in backward order Q[cur].push({0, 0}); md[0][0] = 'X'; //visited while (!Q[cur].empty()) { ans++; //BFS to find the CC for current animal's footprints while (!Q[cur].empty()) { int r = Q[cur].front().first, c = Q[cur].front().second; Q[cur].pop(); int dr = 0, dc = 1; for (int k = 0; k < 4; k++) { //4 directions int nr = r + dr, nc = c + dc; if (nr >= 0 && nr < H && nc >= 0 && nc < W && md[nr][nc] != '.' && md[nr][nc] != 'X') { if (md[nr][nc] == animal[cur]) { //current animial footprint Q[cur].push({nr, nc}); } else { //immediately preceding animial footprint Q[cur^1].push({nr, nc}); //different use // Or you can do: Q[1-cur].emplace_back(nr, nc); } md[nr][nc] = 'X'; //visited } //the following 2 statements for iterating thru 4 directions // you can use: int dr[] = {0, 1, 0, -1}, dc[] = {1, 0, -1, 0}; swap(dr, dc); dc = -dc; } } cur ^= 1; //for immediately preceding animal. or cur = 1 - cur; } return ans; } int main() { ios::sync_with_stdio(false); cin.tie(nullptr); cin.exceptions(cin.failbit); cout << solve() << endl; return 0; }
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