이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include<bits/stdc++.h>
#define pi 3.141592653589793238
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
#define MOD 1000000007
#define INF 999999999999999999
#define pb push_back
#define ff first
#define ss second
#define mt make_tuple
#define ll long long
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
typedef tree<ll, null_type, less_equal<ll>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
const int N = 1e3 + 1;
string s[N];
ll n, m;
char c;
ll curr;
int visited[N][N];
vector<ll> X = {1, -1, 0, 0};
vector<ll> Y = {0, 0, 1, -1};
bool valid(ll x, ll y){
if(x < 0 || x >= n || y < 0 || y >= m){
return false;
}
if(s[x][y] == c || s[x][y] == '.'){
return true;
}
return false;
}
ll cnt = 0;
void dfs(ll a, ll b){
visited[a][b] = curr;
if(s[a][b] != '.'){
cnt++;
}
s[a][b] = '.';
for(int i = 0; i < 4; i++){
ll x, y;
x = a + X[i], y = b + Y[i];
if(!valid(x, y)){
continue;
}
if(curr != visited[x][y])
dfs(x, y);
}
}
int main() {
//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);
fast;
ll T = 1;
//cin >> T;
while (T--) {
cin >> n >> m;
for(int i = 0; i < n; i++){
cin >> s[i];
}
c = s[0][0];
ll ans = 0;
while(1){
cnt = 0;
// for(int i = 0; i < n; i++){
// for(int j = 0; j < m; j++){
// visited[i][j] = false;
// }
// }
curr++;
dfs(0, 0);
if(cnt == 0){
break;
}
ans++;
if(c == 'B'){
c = 'T';
}
else{
c = 'B';
}
}
cout << ans << "\n";
}
return 0;
}
컴파일 시 표준 에러 (stderr) 메시지
zoo.cpp:4: warning: ignoring '#pragma GCC optimization' [-Wunknown-pragmas]
4 | #pragma GCC optimization ("O3")
|
zoo.cpp:5: warning: ignoring '#pragma GCC optimization' [-Wunknown-pragmas]
5 | #pragma GCC optimization ("unroll-loops")
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