백준 21922 학부 연구생 민상 혼내주기

 

오랜만에 구현 문제를 푸니까 엄청 귀찮고 잔실수도 많았다...

모든 에어컨을 큐에 넣고 조건에 맞도록 bfs를 돌리면 된다. 

구현은 언제 물 흐르듯 할 수 있을가...

 

#include <bits/stdc++.h>
#define fastio ios::sync_with_stdio(0), cin.tie(0)
using namespace std;
using ll = long long;
using pii = pair<int, int>;
#define all(v) v.begin(), v.end()

int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};

int arr[2002][2002];
int n, m, ans = 0;
vector<pii> acs;
bool visited[2002][2002][4];

struct state {
    int x, y, d;
    state(int xx, int yy, int dd) {
        x = xx, y = yy, d = dd;
    }
};

void input() {
    cin >> n >> m;
    for(int i = 0; i < n; ++i) {
        for(int j = 0; j < m; ++j) {
            cin >> arr[i][j];
            if(arr[i][j] == 9) acs.push_back({i, j});
        }
    }
}

int main() {
    fastio;
    input();

    queue<state> q;
    for(auto it = acs.begin(); it != acs.end(); ++it) {
        pii ac = *it;
        for(int i = 0; i < 4; ++i) {
            visited[ac.first][ac.second][i] = true;
            q.push(state(ac.first, ac.second, i));
        }
    }

    while(!q.empty()) {
        auto [fx, fy, fd] = q.front(); q.pop();

        // move
        int nx = fx + dx[fd], ny = fy + dy[fd];

        if(nx < 0 || nx >= n || ny < 0 || ny >= m || visited[nx][ny][fd]) continue;
        visited[nx][ny][fd] = true;

        // x directional change
        if(!arr[nx][ny]) q.push(state(nx, ny, fd));
        else if(arr[nx][ny] == 1) {
            if(fd & 1) q.push(state(nx, ny, 4 - fd));
            else q.push(state(nx, ny, fd));
        }
        else if(arr[nx][ny] == 2) {
            if(fd & 1) q.push(state(nx, ny, fd));
            else q.push(state(nx, ny, 2 - fd));
        }
        else if(arr[nx][ny] == 3) q.push(state(nx, ny, 3 - fd));
        else {
            if(fd < 2) q.push(state(nx, ny, 1 - fd));
            else q.push(state(nx, ny, 5 - fd));
        }
    }

    for(int i = 0 ; i < n; ++i) {
        for(int j = 0; j < m; ++j) {
            for(int k = 0; k < 4; ++k) {
                if(visited[i][j][k]) {
                    ans++;
                    break;
                }
            }
        }
    }
    cout << ans;

    return 0;
}