Gidhub BE Developer

# [SW Expert Academy] 1953. 탈주범 검거

2018-05-07
goodGid

## Problem

Problem URL : 탈주범 검거

## [1] Answer Code (18. 10. 13)

``````#include <iostream>
#include <cstring>
#include <vector>
#include <algorithm>
#include <queue>
#define p pair<int,int>
using namespace std;

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

int map[50][50];
int v[50][50]; // v is visit
int row,col;

struct Node{
int x;
int y;
int time;
Node(int a, int b, int c): x(a), y(b), time(c) {};
};

bool inRange(int x, int y){
if( x < 0 || x >= row || y < 0 || y >= col)
return false;
return true;
}

int main() {
ios::sync_with_stdio(0);
cin.tie(0); cout.tie(0);

int TC;
cin >> TC;
for(int tc=1; tc<=TC; tc++){
memset(map,0,sizeof(map));
memset(v, 0, sizeof(v));
int st_x;
int st_y;
int L;
cin >> row >> col >> st_x >> st_y >> L ;

for(int i=0; i<row; i++) for(int j=0; j<col; j++) cin >> map[i][j];

queue<Node> q;
q.push(Node(st_x,st_y,1));

int cnt = 1;
while (! q.empty()) {
int x = q.front().x;
int y = q.front().y;
int time = q.front().time;
q.pop();

if(time == L)
break;
v[x][y] = 1;

for(int i=0; i<4; i++){
int nx = x + dx[i];
int ny = y + dy[i];

// 범위 밖인 경우
if(! inRange(nx, ny))
continue;
int now_value = map[x][y];

if( now_value == 1 && ! v[nx][ny]){
int next_value = map[nx][ny];
if( i == 0 ){ // Right
if( next_value == 1 || next_value == 3
|| next_value == 6 || next_value == 7){
cnt++;
v[nx][ny] = 1;
q.push(Node(nx,ny,time+1));
}
}
else if( i == 1){ // Bottom
if( next_value == 1 || next_value == 2
|| next_value == 4 || next_value == 7){
cnt++;
v[nx][ny] = 1;
q.push(Node(nx,ny,time+1));
}
}
else if( i == 2){ // Left
if( next_value == 1 || next_value == 3
|| next_value == 4 || next_value == 5){
cnt++;
v[nx][ny] = 1;
q.push(Node(nx,ny,time+1));
}
}
else if( i == 3){ // Top
if( next_value == 1 || next_value == 2
|| next_value == 5 || next_value == 6){
cnt++;
v[nx][ny] = 1;
q.push(Node(nx,ny,time+1));
}
}
} // end of now_value = 1

else if( now_value == 2 && ! v[nx][ny]){
int next_value = map[nx][ny];
if( i == 0 ){ // Right
}
else if( i == 1){ // Bottom
if( next_value == 1 || next_value == 2
|| next_value == 4 || next_value == 7){
cnt++;
v[nx][ny] = 1;
q.push(Node(nx,ny,time+1));
}
}
else if( i == 2){ // Left
}
else if( i == 3){ // Top
if( next_value == 1 || next_value == 2
|| next_value == 5 || next_value == 6){
cnt++;
v[nx][ny] = 1;
q.push(Node(nx,ny,time+1));
}
}
} // end of now_value = 2

else if( now_value == 3 && ! v[nx][ny]){
int next_value = map[nx][ny];
if( i == 0 ){ // Right
if( next_value == 1 || next_value == 3
|| next_value == 6 || next_value == 7){
cnt++;
v[nx][ny] = 1;
q.push(Node(nx,ny,time+1));
}
}
else if( i == 1){ // Bottom
}
else if( i == 2){ // Left
if( next_value == 1 || next_value == 3
|| next_value == 4 || next_value == 5){
cnt++;
v[nx][ny] = 1;
q.push(Node(nx,ny,time+1));
}
}
else if( i == 3){ // Top
}
} // end of now_value = 3

else if( now_value == 4 && ! v[nx][ny]){
int next_value = map[nx][ny];
if( i == 0 ){ // Right
if( next_value == 1 || next_value == 3
|| next_value == 6 || next_value == 7){
cnt++;
v[nx][ny] = 1;
q.push(Node(nx,ny,time+1));
}
}
else if( i == 1){ // Bottom
}
else if( i == 2){ // Left
}
else if( i == 3){ // Top
if( next_value == 1 || next_value == 2
|| next_value == 5 || next_value == 6){
cnt++;
v[nx][ny] = 1;
q.push(Node(nx,ny,time+1));
}
}
} // end of now_value = 4

else if( now_value == 5 && ! v[nx][ny]){
int next_value = map[nx][ny];
if( i == 0 ){ // Right
if( next_value == 1 || next_value == 3
|| next_value == 6 || next_value == 7){
cnt++;
v[nx][ny] = 1;
q.push(Node(nx,ny,time+1));
}
}
else if( i == 1){ // Bottom
if( next_value == 1 || next_value == 2
|| next_value == 4 || next_value == 7){
cnt++;
v[nx][ny] = 1;
q.push(Node(nx,ny,time+1));
}
}
else if( i == 2){ // Left
}
else if( i == 3){ // Top
}
} // end of now_value = 5

else if( now_value == 6 && ! v[nx][ny]){
int next_value = map[nx][ny];
if( i == 0 ){ // Right
}
else if( i == 1){ // Bottom
if( next_value == 1 || next_value == 2
|| next_value == 4 || next_value == 7){
cnt++;
v[nx][ny] = 1;
q.push(Node(nx,ny,time+1));
}
}
else if( i == 2){ // Left
if( next_value == 1 || next_value == 3
|| next_value == 4 || next_value == 5){
cnt++;
v[nx][ny] = 1;
q.push(Node(nx,ny,time+1));
}
}
else if( i == 3){ // Top
}
} // end of now_value = 6

else if( now_value == 7 && ! v[nx][ny]){
int next_value = map[nx][ny];
if( i == 0 ){ // Right
}
else if( i == 1){ // Bottom
}
else if( i == 2){ // Left
if( next_value == 1 || next_value == 3
|| next_value == 4 || next_value == 5){
cnt++;
v[nx][ny] = 1;
q.push(Node(nx,ny,time+1));
}
}
else if( i == 3){ // Top
if( next_value == 1 || next_value == 2
|| next_value == 5 || next_value == 6){
cnt++;
v[nx][ny] = 1;
q.push(Node(nx,ny,time+1));
}
}
} // end of now_value = 7
} // end of for i
}
cout << "#" << tc << " " << cnt << endl ;
}
return 0;
}
``````

### Review

• 1번 모양(+)일 때 좌우 조건을 빼먹어서 시간이 걸렸다.

• 1번 조건이 아니였다면 50분 컷이였는데 아쉽다.

• 1번 모양
• 상 : 1 2 5 6
• 하 : 1 2 4 7
• 좌 : 1 3 4 5
• 우 : 1 3 6 7
• 2번 모양
• 상 : 1 2 5 6
• 하 : 1 2 4 7
• 3번 모양
• 좌 : 1 3 4 5
• 우 : 1 3 6 7
• 4번 모양
• 상 : 1 2 5 6
• 우 : 1 3 6 7
• 5번 모양
• 하 : 1 2 4 7
• 우 : 1 3 6 7
• 6번 모양
• 하 : 1 2 4 7
• 좌 : 1 3 4 5
• 7번 모양
• 상 : 1 2 5 6
• 우 : 1 3 6 7

## [2] Answer Code (18. 05. 08)

``````
#include <iostream>
#include <vector>
#include <cstring>
#include <queue>
#define P pair<int,int>
using namespace std;
vector<int> v[7];
int map[50][50];
int visit[50][50];
int dx[] = {0,1,1,1,0,-1,-1,-1};
int dy[] = {1,1,0,-1,-1,-1,0,1};

bool chkShape(int dir, int value){
if(dir == 0 || dir == 7){
if(value == 1 || value == 3 || value == 6 || value == 7){
return true;
} else
return false;
}
else if(dir == 1 || dir == 2 || dir == 3){
if(value == 1 || value == 2 || value == 4 || value == 7){
return true;
} else
return false;
}
else if(dir == 4 || dir == 5){
if(value == 1 || value == 3 || value == 4 || value == 5){
return true;
} else
return false;
}
else { // 6
if(value == 1 || value == 2 || value == 5 || value == 6){
return true;
} else
return false;
}
}

vector<int> getDirection(int dir){
vector<int> v;
if(dir == 1){
v.push_back(0); v.push_back(2); v.push_back(4); v.push_back(6);
}
else if( dir == 2){
v.push_back(2); v.push_back(6);
}
else if( dir == 3){
v.push_back(0); v.push_back(4);
}
else if( dir == 4){
v.push_back(0); v.push_back(6);
}
else if( dir == 5 ){
v.push_back(0); v.push_back(2);
}
else if( dir == 6){
v.push_back(2); v.push_back(4);
}
else if( dir == 7){
v.push_back(4); v.push_back(6);
}
return v;
}

int main(){
int tc;
cin >> tc;

for(int st=1; st<=tc; st++){
memset(map,0,sizeof(map));
memset(visit,0,sizeof(visit));

vector<int> v;
queue<P> q;

int n,m,X,Y,time;
cin >> n >> m >> X >> Y >> time;

for(int i=0; i<n; i++){
for(int j=0; j<m; j++){
scanf("%d",&map[i][j]);
}
}

q.push({X,Y});
visit[X][Y] = 1;
int cnt = 0;
for(int i=0; i<time; i++){
int size = (int)q.size();

for(int k=0; k<size; k++){
int x = q.front().first;
int y = q.front().second;
q.pop();
cnt ++ ;

vector<int> _v= getDirection(map[x][y]);
int _v_size = (int)_v.size();
for(int c=0; c<_v_size; c++){
int nx = x + dx[_v[c]];
int ny = y + dy[_v[c]];
if( nx < 0 || nx > n-1 || ny < 0 || ny > m-1) continue;
if(! visit[nx][ny] && chkShape(_v[c],map[nx][ny])){
visit[nx][ny] = 1;
q.push({nx,ny});
}
} // End of y for Loop
} // End of k for Loop
} // End of i for Loop
printf("#%d %d\n",st,cnt);
}

return 0;
}
``````

### Review

• 파이프 모양에 따른 호환성 체크 조건을 빼먹고 코딩을 시작했다.

• 모양에 따른 호환성 체크를 어떻게 해주면 좋을까?

• 2시간 30분 걸림

• chkShape()를 만드는데 굉장히 오랜 시간이 걸렸다.

• getDirection()를 만드는데 오랜 시간이 걸리지 않았지만 실수를 했다.