[C++] 백준 8972 미친 아두이노

시뮬레이션 + 구현 문제  

 

이전 배열의 상태를 저장해 주기만 하면 쉽게 풀수있다.

 

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <cmath>
 
using namespace std;
 
string arr;
string str;
vector<int> moving;
vector<pair<int,int> > mad;
int dx[] = {1,1,1,0,0,0,-1,-1,-1};
int dy[] = {-1,0,1,-1,0,1,-1,0,1};
int n,m,x,y;
int cnt = 0;
 
// 1 == 아두이노, 2 == 미친 아두이노 
int v[101][101];
int rem[101][101];
 
void input() {
    cin >> n >> m;
    for(int i=0; i<n; i++) {
        cin >> arr;
        for(int j=0; j<arr.length(); j++) {                                                                        
            if(arr[j] == 'I') {
                 v[i][j] = 1;
                x = i;
                y = j; 
            }
            if(arr[j] == 'R') {
                v[i][j] = 2;
                mad.push_back({i,j});
            }
        }
    }
    
    cin >> str;
    for(int i=0; i<str.length(); i++) {
        moving.push_back(str[i] - '0');
    }
}
 
void print() {
    for(int i=0; i<n; i++) {
        for(int j=0; j<m; j++) {
            if(v[i][j] == 1) { cout << "I"; }
            else if(v[i][j] == 2) { cout << "R"; }
            else { cout << "."; }
        }
        cout << '\n';
    }
}
 
void copy() {
    for(int i=0; i<n; i++) {
        for(int j=0; j<m; j++) {
            if(rem[i][j] == 1 || rem[i][j] == 2) {
                v[i][j] = rem[i][j];
            }
            else {
                v[i][j] = 0;
            }
            rem[i][j] = 0;
            
        }
    }
}
 
void solve() {
    // 종수 움직임 
    for(int i : moving) {
        cnt++;
        int kx = dx[i-1] + x;
        int ky = dy[i-1] + y;
        
        if(v[kx][ky] == 2) {
            cout << "kraj " << cnt;
            return;
        }
        
        
        rem[kx][ky] = 1;
        x = kx;
        y = ky;
        
        
        for(int j=0; j<n; j++) {
            for(int k=0; k<m; k++) {
                if(v[j][k] == 2) {
                    int min_tmp = 99999999;
                    int d;
                    
                    for(int q=0; q<9; q++) {
                        int nx = j + dx[q];
                        int ny = k + dy[q];
                        int tmp;
                        tmp = fabs(x - nx) + fabs(y - ny);
                        if(tmp < min_tmp) {
                            min_tmp = tmp;
                            d = q;
                        }
                    }
                    
                    int nx = dx[d] + j;
                    int ny = dy[d] + k;
                    if(x == nx && y == ny) {
                        cout << "kraj " << cnt;
                        return;
                    }
                    rem[nx][ny] += 2;
                }
            }
        }
        
        copy();
    }
    
    print();
}
 
int main() {
    input();
    solve();
    
}