제출 #414156

#제출 시각아이디문제언어결과실행 시간메모리
414156600MihneaCultivation (JOI17_cultivation)C++17
30 / 100
570 ms740 KiB
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
#define int ll

ll max(ll a, signed b) {
  if (a > b) return a;
  return b;
}


const int N = 300 + 7;
const int INF = (int) 1e18 + 7;

int dimx, dimy, n;

struct point {
  int x;
  int y;
};

point v[N];
vector<int> cover[N];

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

  ///freopen ("input", "r", stdin);

  cin >> dimx >> dimy >> n;

  for (int i = 1; i <= n; i++) {
    cin >> v[i].x >> v[i].y;
  }

  assert(dimx <= 40);

  int sol = INF;

  for (int a = 0; a <= dimx; a++) {
    for (int b = 0; b <= dimx; b++) { /// fix north, south

      for (int i = 1; i <= dimx; i++) {
        cover[i].clear();
      }
      for (int i = 1; i <= n; i++) {
        int l = max(1, v[i].x - a), r = min(dimx, v[i].x + b);
        for (int j = l; j <= r; j++) {
          cover[j].push_back(i);
        }
      }

      bool ok = 1;
      for (int i = 1; i <= dimx; i++) {
        if (cover[i].empty()) {
          ok = 0;
          break;
        }
      }

      if (!ok) {
        continue;
      }

      /// now it's a pretty easy task

      /// For each row I have some restrictions
      int r1 = 0, r2 = 0, rsum = 0;

      for (int i = 1; i <= dimx; i++) {
        vector<int> pts;
        for (auto &j : cover[i]) {
          pts.push_back(v[j].y);
        }

        sort(pts.begin(), pts.end());

        r1 = max(r1, pts[0] - 1);
        r2 = max(r2, dimy - pts.back());

        for (int j = 1; j < (int) pts.size(); j++) {
          rsum = max(rsum, pts[j] - pts[j - 1] - 1);
        }
      }
      rsum = max(rsum, r1 + r2);
      sol = min(sol, a + b + rsum);
    }
  }

  cout << sol << "\n";
  return 0;
}
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