Table of contents

• Problem
  • Input
  • Output
• Time complexity
• Solution

# Problem

https://atcoder.jp/contests/arc031/tasks/arc031_2

# Input

$c_1 c_2 ... c_{10}$
$c_{11} c_{12} ... c_{20}$
...
$c_{91} c_{92} ... c_{100}$.

• $c_i$ - Either x or o.

# Output

Report if all the o are connected as 4-neighbor (up, down, left or right).
You can flip a x to o.

# Time complexity

$O(H W)$ for bit exhaustive search.
And $O(H W)$ for DFS for each step in bit exhaustive search.

$O((H W)^2)$

# Solution

// C++ 14
#include <iostream>
#include <string>
#include <vector>
#include <list>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <unordered_map>
#include <math.h>

#define ll long long
#define Int int
#define loop(x, start, end) for(Int x = start; x < end; x++)
#define loopdown(x, start, end) for(int x = start; x > end; x--)
#define rep(n) for(int x = 0; x < n; x++)
#define span(a,x,y) a.begin()+x,a.begin()+y
#define span_all(a) a.begin(),a.end()
#define len(x) (x.size())
#define last(x) (*(x.end()-1))

using namespace std;

#define MAX_N 10

Int H, W;
vector<bool> C(MAX_N*MAX_N, true);

void input() {
  char x;
  H = W = 10;
  loop(h,0,H) {
    loop(w,0,W) {
      cin >> x;
      if (x == 'x') C[h*W + w] = false;
    }
  }
}

void dfs(Int x, Int y, Int depth, vector<bool> &c) {
  c[y*W+x] = false;
  loop(dy,-1,2) {
    loop(dx,-1,2) {
      if (dy != 0 && dx != 0) continue;
      Int x_ = x + dx, y_ = y + dy;
      if (x_ < 0 || W <= x_ || y_ < 0 || H <= y_) continue;
      if (!c[y_*W + x_]) continue;
      dfs(x_, y_, depth+1, c);
    }
  }
}

bool bit(Int x, Int y, vector<bool> c) {
  c[y*W+x] = true;
  dfs(x, y, 0, c);
  loop(h,0,H) {
    loop(w,0,W) {
      if (c[h*W+w]) return false;
    }
  }

  return true;
}

void solve() {
  loop(h,0,H) {
    loop(w,0,W) {
      if (bit(w, h, C)) {
        cout << "YES" << endl;
        return;
      }
    }
  }

  cout << "NO" << endl;
}

int main(void) {
  input();
  solve();
  return 0;
}