프로그래머스 거리두기확인하기 c++ // bfs, check함수 활용방법, 문자열을 배열처럼 활용하는방법

https://school.programmers.co.kr/learn/courses/30/lessons/81302

프로그래머스

1. 문자열을 배열처럼 활용하는방법

auto place : places
후
place[i][j]로 접근하면 된다.

places -> [POOOX, PPPPP..], [PPAP, ...] [...]
place -> [POOOX, PPPPP, ...]
place[i] - > POOOX
place[i][0]=='P'

 

2. check함수 활용방법

1. check로 의사코드짜고
2. 이후에 구체적으로 구현하도록 하자

for(auto place : places){
        if(check(place)) answer.push_back(1); //지킴
        else answer.push_back(0); //안지킴
    }

 

3. 전체코드

#include <bits/stdc++.h>

using namespace std;

int vis[5][5];
int dy[]={1,0,-1,0};
int dx[]={0,1,0,-1};
vector<int> answer;
struct Point{
    int y;
    int x;
    int dist;
};
bool bfs(vector<string>& place,int y, int x){
    queue<Point> q;
    q.push({y,x,0});
    bool v[5][5]={false};
    v[y][x]=1;
    
    while(q.size()){
        Point curr=q.front(); q.pop();
        if(curr.dist>2) continue;
        //거리 2이하 and 초기위치아님 and 사람을만낫다 -> 거리두기 실패 false
        if(curr.dist!=0 and place[curr.y][curr.x]=='P') return false;
        
        for(int i=0;i<4;++i){
            int ny=curr.y+dy[i];
            int nx=curr.x+dx[i]; 
            
            if(ny<0 ||ny>=5 || nx<0||nx>=5) continue;
            if(v[ny][nx]) continue;
            if(place[ny][nx]=='X')continue;
            
            v[ny][nx]=1;
            q.push({ny,nx,curr.dist+1});
        }
    }
    return true;
    
}
bool check(vector<string> & place){
    for(int i=0;i<5;++i){
        for(int j=0;j<5;++j){
           if(place[i][j]=='P'){
               if(bfs(place,i,j)==false) return false;
           }
        }
    }    
    return true;
}
vector<int> solution(vector<vector<string>> places) {

    for(auto place : places){
        if(check(place)) answer.push_back(1); //지킴
        else answer.push_back(0); //안지킴
    }
    return answer;
}