제출 #470762

#제출 시각아이디문제언어결과실행 시간메모리
470762Yazan_AlattarMecho (IOI09_mecho)C++14
4 / 100
40 ms4548 KiB
#include <iostream> #include <fstream> #include <vector> #include <cstring> #include <algorithm> #include <set> #include <map> #include <queue> #include <utility> #include <cmath> #include <numeric> using namespace std; typedef long long ll; #define F first #define S second #define pb push_back #define endl "\n" #define all(x) x.begin(), x.end() const int M = 1007; const ll inf = 1e18; const ll mod = 1e9 + 7; const double pi = acos(-1); const int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1}; queue < pair <int,int> > q; pair <int,int> st, en, p[M][M]; char a[M][M]; int n, s, d[M][M], dist[M][M]; int main() { ios::sync_with_stdio(0);cin.tie(0);cout.tie(0); cin >> n >> s; for(int i = 1; i <= n; ++i){ for(int j = 1; j <= n; ++j){ cin >> a[i][j]; if(a[i][j] == 'H') q.push({i, j}), d[i][j] = 1; if(a[i][j] == 'D') en = {i, j}; if(a[i][j] == 'M') st = {i, j}; } } while(!q.empty()){ int x = q.front().S; int y = q.front().F; q.pop(); for(int i = 0; i < 4; ++i){ int nx = x + dx[i]; int ny = y + dy[i]; if(a[ny][nx] != 'G' || d[ny][nx] || ny < 1 || nx < 1 || ny > n || nx > n) continue; d[ny][nx] = d[y][x] + 1; q.push({ny, nx}); } } // for(int i = 1; i <= n; ++i){ // for(int j = 1; j <= n; ++j){ // cout << d[i][j] << " "; // } // cout << endl; // } q.push(st); dist[st.F][st.S] = 1; while(!q.empty()){ int x = q.front().S; int y = q.front().F; q.pop(); for(int i = 0; i < 4; ++i){ int nx = x + dx[i]; int ny = y + dy[i]; if(a[ny][nx] != 'G' || (dist[y][x]) / s >= d[y][x] - 1 || dist[ny][nx] || ny < 1 || nx < 1 || ny > n || nx > n) continue; p[ny][nx] = {y, x}; dist[ny][nx] = dist[y][x] + 1; q.push({ny, nx}); } } // cout << endl; // for(int i = 1; i <= n; ++i){ // for(int j = 1; j <= n; ++j) cout << dist[i][j] << " "; // cout << endl; // } if(!dist[en.F][en.S]) return cout << -1 << endl, 0; int ans = 1e9; int i = en.F, j = en.S; while(i != st.F || j != st.S){ ans = min(ans, (d[i][j] - 1) - ((dist[i][j] - 2) / s + 1)); // cout << i << " " << j << " " << (d[i][j] - 1) - ((dist[i][j] - 2) / s + 1) << " " << d[i][j] - 1 << " " << (dist[i][j] - 1) / s << endl; int ni = p[i][j].F, nj = p[i][j].S; i = ni; j = nj; } cout << ans << endl; return 0; } // Don't forget special cases. (n = 1?) // Look for the constraints. (Runtime array? overflow?)
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