| # | Time | Username | Problem | Language | Result | Execution time | Memory |
|---|---|---|---|---|---|---|---|
| 498315 | chenwz | Mecho (IOI09_mecho) | C++11 | 0 ms | 0 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
// IOI2009 - Mecho
#include <bits/stdc++.h>
using namespace std;
const int NN = 2000 + 4, DX[4] = {1, -1, 0, 0}, DY[4] = {0, 0, 1, -1};
#define _for(i, a, b) for (int i = (a); i < (int)(b); ++i)
string G[NN];
int N, S, D[NN][NN];
bool Vis[NN][NN];
struct Point {
int x, y;
};
bool valid(int x, int y) {
return 0 <= x && x < N && 0 <= y && y < N;
// nx < 0 || nx >= N || ny < 0 || ny >= N
}
bool check(const Point& st, const Point& Home,
int delay) { // 小熊delay再出发够不?
if (delay * S >= D[st.x][st.y]) return false;
queue<pair<int, Point> > Q; // {当前时间，当前位置}
memset(Vis, 0, sizeof(Vis)), Q.push({delay * S, st}), Vis[st.x][st.y] = true;
while (!Q.empty()) {
auto p = Q.front();
Q.pop();
int d = p.first, x = p.second.x, y = p.second.y;
// if (G[x][y] == 'D')
if (x == Home.x && t == Home.y) return true; // 到家
_for(i, 0, 4) {
int nx = x + DX[i], ny = y + DY[i], &nd = D[nx][ny];
if (valid(nx, ny) && G[nx][ny] != 'T' && d + 1 < D[nx][ny] &&
!Vis[nx][ny])
// if (nx < 0 || nx >= N || ny < 0 || ny >= N || G[nx][ny] == 'T' ||
// (d + 1) >= D[nx][ny] || Vis[nx][ny])
// continue;
Q.push({d + 1, {nx, ny}}), Vis[nx][ny] = true;
}
}
return false;
}
int main() {
ios::sync_with_stdio(false), cin.tie(0);
cin >> N >> S;
queue<Point> Q;
Point Honey, Home;
memset(D, -1, sizeof(D));
_for(i, 0, N) {
cin >> G[i];
_for(j, 0, N) {
char& c = G[i][j];
if (c == 'H') Q.push({i, j}), D[i][j] = 0;
if (c == 'M') Honey = {i, j}, c = 'G'; // 初始小熊位置也改成草
if (c == 'D') Home = {i, j}; // 家的位置
}
}
while (!Q.empty()) { // BFS计算每个点到最近的蜜蜂的距离
Point p = Q.front();
Q.pop();
_for(i, 0, 4) {
int nx = p.x + DX[i], ny = p.y + DY[i], &nd = D[nx][ny];
if (valid(nx, ny) && G[nx][ny] == 'G' && nd == -1)
// if (!valid(nx, ny) || G[nx][ny] != 'G' || D[nx][ny] != -1) continue;
nd = D[p.x][p.y] + S, Q.push({nx, ny});
}
}
D[Home.x][Home.y] = N * N * S;
int L = -1, R = 2 * N * N;
while (L + 1 < R) {
int mid = (L + R) / 2; // 推迟mid出发够不够
(check(Honey, mid) ? L : R) = mid;
}
cout << L << endl;
return 0;
}
// 2021-12-20 Mecho (IOI09_mecho) C++11 100/100 180 ms 21600 KB