[삼성 SW 역량 테스트] 감시
in Coding Test on 삼성 역량테스트
시뮬레이션과 DFS의 혼종.
때로는 하드코딩이 시간을 더 단축시킬 때가 있다.
Code
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
int N, M;
int ans = 987654321;
int map[8][8];
vector<pair<int, int>> v;
const int dy[] = { -1,0,1,0 };
const int dx[] = { 0,1,0,-1 };
void move(int dir, int y, int x) {
switch (dir) {
case 0:
for (int i = y - 1; i >= 0; i--) {
if (map[i][x] == 6) break;
if (map[i][x] == 0) map[i][x] = -1;
}
break;
case 1:
for (int j = x + 1; j < M; j++) {
if (map[y][j] == 6) break;
if (map[y][j] == 0) map[y][j] = -1;
}
break;
case 2:
for (int i = y + 1; i < N; i++) {
if (map[i][x] == 6) break;
if (map[i][x] == 0) map[i][x] = -1;
}
break;
case 3:
for (int j = x - 1; j >= 0; j--) {
if (map[y][j] == 6) break;
if (map[y][j] == 0) map[y][j] = -1;
}
break;
}
}
void copy(int a[8][8], int b[8][8]) {
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
a[i][j] = b[i][j];
}
}
}
void dfs(int cnt) {
if (cnt == v.size()) {
int ret = 0;
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
if (map[i][j] == 0)
ret++;
}
}
ans = min(ans, ret);
if (ans == 0) {
printf("%d\n", ans);
exit(0);
}
return;
}
int temp[8][8] = { 0, };
copy(temp, map);
int y = v[cnt].first;
int x = v[cnt].second;
int number = map[y][x];
switch (number) {
case 1:
for (int dir = 0; dir < 4; dir++) {
move(dir, y, x);
dfs(cnt + 1);
copy(map,temp);
}
break;
case 2:
for (int dir = 0; dir < 2; dir++) {
move(dir, y, x);
move(dir + 2, y, x);
dfs(cnt + 1);
copy(map, temp);
}
break;
case 3:
for (int dir = 0; dir < 4; dir++) {
move(dir, y, x);
move((dir + 1) % 4, y, x);
dfs(cnt + 1);
copy(map, temp);
}
break;
case 4:
for (int dir = 0; dir < 4; dir++) {
move(dir, y, x);
move((dir + 1) % 4, y, x);
move((dir + 2) % 4, y, x);
dfs(cnt + 1);
copy(map, temp);
}
break;
case 5:
for (int dir = 0; dir < 4; dir++) {
move(dir, y, x);
}
dfs(cnt + 1);
copy(map, temp);
break;
}
return;
}
int main() {
scanf("%d %d", &N, &M);
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
scanf("%d", &map[i][j]);
if (map[i][j] != 0 && map[i][j] != 6) {
v.push_back({ i,j });
}
}
}
dfs(0);
printf("%d\n", ans);
return 0;
}