SRM 636 div1 easy ChocolateDividingEasy

解法

二次元累積和を計算しておくと矩形の範囲の値の総和をO(1)で計算できる。あとは切る位置を全通り試すO(H^2W^2)をすればよい。

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<ll> VL;
typedef vector<VL> VVL;
typedef pair<int, int> PII;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
#define IN(a, b, x) (a<=x&&x<b)
#define MP make_pair
#define PB push_back
const int INF = (1LL<<30);
const ll LLINF = (1LL<<60);
const double PI = 3.14159265359;
const double EPS = 1e-12;
const int MOD = 1000000007;
//#define int ll

template <typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template <typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }

int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};

int d[55][55] = {};
class ChocolateDividingEasy {
   public:
   int findBest(vector <string> choco)
  {
    int h = choco.size(), w = choco[0].size();
    REP(i, h) REP(j, w) d[i+1][j+1] = choco[i][j]-'0';
    FOR(i, 1, h+1) FOR(j, 1, w+1) d[i][j] += d[i-1][j] + d[i][j-1] - d[i-1][j-1];

    int ret = 0;
    FOR(i, 1, h) FOR(j, i+1, h) FOR(k, 1, w) FOR(l, k+1, w) {
      int v1 = d[i][k],
          v2 = d[i][l] - v1,
          v3 = d[j][k] - v1,
          v4 = d[j][l] - v2 - v3 - v1,
          v5 = d[i][w] - v1 - v2,
          v6 = d[j][w] - v1 - v2 - v3 - v4 - v5,
          v7 = d[h][k] - v1 - v3,
          v8 = d[h][l] - v1 - v2 - v3 - v4 - v7,
          v9 = d[h][w] -v1-v2-v3-v4-v5-v6-v7-v8;
      chmax(ret, min({v1, v2, v3, v4, v5, v6, v7, v8, v9}));
    }

    return ret;
  }
};