結果

問題 No.460 裏表ちわーわ
ユーザー koyumeishikoyumeishi
提出日時 2016-12-11 00:50:03
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
RE  
実行時間 -
コード長 5,635 bytes
コンパイル時間 1,440 ms
コンパイル使用メモリ 120,004 KB
実行使用メモリ 6,824 KB
最終ジャッジ日時 2024-11-29 02:41:42
合計ジャッジ時間 3,451 ms
ジャッジサーバーID
(参考情報)
judge2 / judge4
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 2 ms
6,816 KB
testcase_01 AC 2 ms
6,820 KB
testcase_02 AC 2 ms
6,816 KB
testcase_03 AC 2 ms
6,820 KB
testcase_04 AC 2 ms
6,816 KB
testcase_05 RE -
testcase_06 RE -
testcase_07 RE -
testcase_08 AC 2 ms
6,820 KB
testcase_09 AC 2 ms
6,816 KB
testcase_10 AC 2 ms
6,820 KB
testcase_11 RE -
testcase_12 RE -
testcase_13 WA -
testcase_14 WA -
testcase_15 WA -
testcase_16 RE -
testcase_17 WA -
testcase_18 WA -
testcase_19 WA -
testcase_20 RE -
testcase_21 WA -
testcase_22 WA -
testcase_23 WA -
testcase_24 AC 2 ms
6,816 KB
testcase_25 AC 2 ms
6,820 KB
testcase_26 AC 2 ms
6,816 KB
testcase_27 AC 2 ms
6,816 KB
権限があれば一括ダウンロードができます

ソースコード

diff #

#include <iostream>
#include <vector>
#include <cstdio>
#include <sstream>
#include <map>
#include <string>
#include <algorithm>
#include <queue>
#include <cmath>
#include <functional>
#include <set>
#include <ctime>
#include <random>
#include <chrono>
#include <cassert>
#include <tuple>
#include <utility>
using namespace std;

namespace {
  using Integer = long long; //__int128;
  template<class T, class S> istream& operator >> (istream& is, pair<T,S>& p){return is >> p.first >> p.second;}
  template<class T> istream& operator >> (istream& is, vector<T>& vec){for(T& val: vec) is >> val; return is;}
  template<class T> istream& operator ,  (istream& is, T& val){ return is >> val;}
  template<class T, class S> ostream& operator << (ostream& os, const pair<T,S>& p){return os << p.first << " " << p.second;}
  template<class T> ostream& operator << (ostream& os, const vector<T>& vec){for(size_t i=0; i<vec.size(); i++) os << vec[i] << (i==vec.size()-1?"":" "); return os;}
  template<class T> ostream& operator ,  (ostream& os, const T& val){ return os << " " << val;}

  template<class H> void print(const H& head){ cout << head; }
  template<class H, class ... T> void print(const H& head, const T& ... tail){ cout << head << " "; print(tail...); }
  template<class ... T> void println(const T& ... values){ print(values...); cout << endl; }

  template<class H> void eprint(const H& head){ cerr << head; }
  template<class H, class ... T> void eprint(const H& head, const T& ... tail){ cerr << head << " "; eprint(tail...); }
  template<class ... T> void eprintln(const T& ... values){ eprint(values...); cerr << endl; }

  class range{ Integer start_, end_, step_; public: struct range_iterator{ Integer val, step_; range_iterator(Integer v, Integer step) : val(v), step_(step) {} Integer operator * (){return val;} void operator ++ (){val += step_;} bool operator != (range_iterator& x){return step_ > 0 ? val < x.val : val > x.val;} }; range(Integer len) : start_(0), end_(len), step_(1) {} range(Integer start, Integer end) : start_(start), end_(end), step_(1) {} range(Integer start, Integer end, Integer step) : start_(start), end_(end), step_(step) {} range_iterator begin(){ return range_iterator(start_, step_); } range_iterator   end(){ return range_iterator(  end_, step_); } };

  inline string operator "" _s (const char* str, size_t size){ return move(string(str)); }
  constexpr Integer my_pow(Integer x, Integer k, Integer z=1){return k==0 ? z : k==1 ? z*x : (k&1) ? my_pow(x*x,k>>1,z*x) : my_pow(x*x,k>>1,z);}
  constexpr Integer my_pow_mod(Integer x, Integer k, Integer M, Integer z=1){return k==0 ? z%M : k==1 ? z*x%M : (k&1) ? my_pow_mod(x*x%M,k>>1,M,z*x%M) : my_pow_mod(x*x%M,k>>1,M,z);}
  constexpr unsigned long long operator "" _ten (unsigned long long value){ return my_pow(10,value); }

  inline int k_bit(Integer x, int k){return (x>>k)&1;} //0-indexed

  mt19937 mt(chrono::duration_cast<chrono::nanoseconds>(chrono::steady_clock::now().time_since_epoch()).count());

  template<class T> string join(const vector<T>& v, const string& sep){ stringstream ss; for(size_t i=0; i<v.size(); i++){ if(i>0) ss << sep; ss << v[i]; } return ss.str(); }

  inline string operator * (string s, int k){ string ret; while(k){ if(k&1) ret += s; s += s; k >>= 1; } return ret; }
}
constexpr long long mod = 9_ten + 7;

// A[n*p] * X[p*m] = B[n*m]
template<class T = long long>
vector<T> gaussian_elimination_binary(vector<T> A, vector<T> B, int n, int p, int m){
  vector<T> M(n);
  for(int i=0; i<n; i++){
    M[i] = A[i] << m;
    M[i] |= B[i];
  }

  int rank = 0;
  vector<int> left_most(n,-1);
  for(int row = 0, col = p+m-1; row<n && col>=m; col--){
    if( ((M[row] >> col) & T(1)) == T(0)){
      int pivot = row;
      for(int row__ = row; row__<n; row__++){
        if( ((M[row__] >> col) & T(1)) != T(0) ){
          pivot = row__;
          break;
        }
      }

      if(pivot == row) continue;
      swap(M[row], M[pivot]);
    }

    rank++;
    left_most[row] = col;

    for(int row__=row+1; row__<n; row__++){
      if( ((M[row__] >> col) & T(1)) != T(0) ) M[row__] ^= M[row];
    }

    row++;
  }

  // T = long long 
  //T mask_B = ((T)1<<m) - 1;
  //T mask_A = (((T)1<<(m+p)) - 1) ^ mask_B;

  // T = bitset
  T mask_B( string(m, '1') );
  T mask_A( string(m+p, '1') );
  mask_A ^= mask_B;

  bool has_solution = true;
  for(int row=n-1; row>=0; row--){
    if( (M[row] & mask_B) != T(0) && (M[row] & mask_A) == T(0) ) has_solution = false;
    if(left_most[row] == -1) continue;
    int col = left_most[row];
    for(int row__ = row-1; row__>=0; row__--){
      if( ((M[row__] >> col) & T(1)) != T(0) ) M[row__] ^= M[row];
    }
  }

  return has_solution? M : vector<T>(0);
}

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

#include <bitset>
using bits = bitset<300>;

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

  vector<vector<int>> v(n, vector<int>(m));
  cin >> v;

  vector<bits> B(n*m);
  for(int i=0; i<n; i++){
    for(int j=0; j<m; j++){
      if( v[i][j] ) B[i*m+j].set(0, 1);
    }
  }

  //eprintln("B\n", join(B,"\n"));

  vector<bits> A(n*m);
  for(int i=0; i<n; i++){
    for(int j=0; j<m; j++){
      A[i*m + j].set(i*m+j, 1);
      for(int k=0; k<8; k++){
        int y = i+dy[k];
        int x = j+dx[k];
        if(y<0 || y>=n || x<0 || x>=m) continue;
        A[i*m+j].set(y*m+x, 1);
      }
    }
  }

  //eprintln("A\n", join(A,"\n"));

  auto res = gaussian_elimination_binary<bits>(A,B, n*m, n*m, 1);

  //eprintln( join(res, "\n"));

  int ans = 0;
  for(int i=0; i<m*n; i++){
    ans += res[i][0];
  }

  println(ans);


  return 0;
}
0