AOJ0560 Planetary Exploration

問題文
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0560

2次元imos.

  1. f[a-1][b-1]
  2. f[a-1][d]
  3. f[c][b-1]
  4. f[c][d]

1 2
3 4

とすると,で求めることが出来る.

Code

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#include <iostream>
#include <vector>
#include <string>
#include <cstring>
#include <algorithm>
#include <sstream>
#include <map>
#include <set>

#define REP(i,k,n) for(int i=k;i<n;i++)
#define rep(i,n) for(int i=0;i<n;i++)
#define INF 1<<30
#define pb push_back
#define mp make_pair

using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int J[1005][1005], O[1005][1005], I[1005][1005];

int main() {
  int m, n, k;
  cin >> m >> n >> k;

  vector<string> v(m);
  rep(i, m) cin >> v[i];

  memset(J, 0, sizeof(J));
  memset(O, 0, sizeof(O));
  memset(I, 0, sizeof(I));

  rep(i, m) {
      rep(j, n) {
          if(v[i][j] == 'J') J[i][j]++;
          if(v[i][j] == 'O') O[i][j]++;
          if(v[i][j] == 'I') I[i][j]++;
      }
  }

  rep(i, m) {
      REP(j, 1, n) {
          J[i][j] += J[i][j-1];
          O[i][j] += O[i][j-1];
          I[i][j] += I[i][j-1];
      }
  }

  rep(i, n) {
      REP(j, 1, m) {
          J[j][i] += J[j-1][i];
          O[j][i] += O[j-1][i];
          I[j][i] += I[j-1][i];
      }
  }

  rep(i, k) {
      int a, b, c, d;
      cin >> a >> b >> c >> d;

      a--;
      b--;
      c--;
      d--;

      cout << J[c][d] - (a-1 >= 0 ? J[a-1][d] : 0) - (b-1 >= 0 ? J[c][b-1] : 0) + (a-1 >= 0 && b -1 >= 0 ? J[a-1][b-1] : 0) << " ";
      cout << O[c][d] - (a-1 >= 0 ? O[a-1][d] : 0) - (b-1 >= 0 ? O[c][b-1] : 0) + (a-1 >= 0 && b -1 >= 0 ? O[a-1][b-1] : 0) << " ";
      cout << I[c][d] - (a-1 >= 0 ? I[a-1][d] : 0) - (b-1 >= 0 ? I[c][b-1] : 0) + (a-1 >= 0 && b -1 >= 0 ? I[a-1][b-1] : 0) << endl;
  }

  return 0;
}

ベースを

Feb 12th, 2016