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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <functional>
using namespace __gnu_pbds;
using namespace std;
typedef tree<int, null_type, less<int>, rb_tree_tag,
tree_order_statistics_node_update>
ordered_set;
//Dijkstra->set
//set.find_by_order(x) x-position value
//set.order_of_key(x) number of strictly less elements don't need *set.??
#define N 100005
#define wr cout << "Continue debugging\n";
#define all(x) (x).begin(), (x).end()
#define ll long long int
#define pii pair <int, int>
#define pb push_back
#define ff first
#define ss second
int X[] = {-1, 1, 0, 0};
int Y[] = {0, 0, -1, 1};
int main ()
{
ios::sync_with_stdio(false);
cin.tie(0);
int n, m;
cin >> n >> m;
vector<vector<char>> c(n+1, vector<char>(m+1));
for (int i = 1; i <= n; i++){
for (int j = 1; j <= m; j++) cin >> c[i][j];
}
int ans = 0;
vector<vector<bool>> vis(n+1, vector<bool>(m+1, 0));
priority_queue<pair<int,pii>> q; q.push({-1, {1, 1}});
while(!q.empty()){
int x = q.top().ss.ff, y = q.top().ss.ss, w = q.top().ff;
q.pop();
if (vis[x][y]) continue;
vis[x][y] = 1;
ans = max(ans, -w);
for (int i = 0; i < 4; i++){
int a = x+X[i], b = y+Y[i];
if (a > 0 and b > 0 and a <= n and b <= m and !vis[a][b] and c[a][b] != '.'){
if (c[x][y] != c[a][b]) q.push({-(w+1), {a, b}});
else q.push({-w, {a, b}});
}
}
}
cout << ans << '\n';
}
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