Luogu1_4 cqoi2014 harmony matrix exclusive or Gaussian Element

Portal

$ N, m $ is given. Create a non-full $0 $ matrix of $ n \ times M $. All the grids meet the following conditions: it is an even sum of the weights of the top, bottom, and left grids. $ N, m \ Leq 40 $

 

You can list the differences or equations of $ n \ times M $ and $ n \ times M $ equations based on the conditions in the question. (The equations mean that all positions are $1 $ or $0 $, the answers in the rightmost column need to be generated in the sense of $ mod \, 2 $ through equations of different or mutual elimination ).

In theory, when the number of the same element and the number of equations is the same, each element has a unique solution. However, in the case of a solution, some of the equations are linearly related, which means they are eliminated to the end, some rows are changed to $0 $ (if not clearly, you can create a $3 \ times 3 $ and $4 \ times 4 $ table like $ vegetable chicken $ me ). We can set all $0 $ yuan (also called free yuan) in a row to $1 $, because they have no relationship with the final solution of the equation, then, we can launch the above step by step.

Because the complexity is $1600 ^ 3 $, the normal Gaussian elimination speed is very slow, so you can use fairy $ STL \, bitset $ to optimize

 1 #include<bits/stdc++.h> 2 using namespace std; 3  4 inline int read(){ 5     int a = 0; 6     bool f = 0; 7     char c = getchar(); 8     while(c != EOF && !isdigit(c)){ 9         if(c == ‘-‘)10             f = 1;11         c = getchar();12     }13     while(c != EOF && isdigit(c)){14         a = (a << 3) + (a << 1) + (c ^ ‘0‘);15         c = getchar();16     }17     return f ? -a : a;18 }19 20 const int dir[5][2] = {0,1,0,-1,1,0,-1,0,0,0};21 bitset < 1600 > gauss[1600] , ans;22 23 int main(){24 #ifdef LG25     freopen("3164.in" , "r" , stdin);26     freopen("3164.out" , "w" , stdout);27 #endif28     int M , N;29     cin >> M >> N;30     for(int i = 0 ; i < M ; i++)31         for(int j = 0 ; j < N ; j++)32             for(int k = 0 ; k < 5 ; k++)33                 if(i + dir[k][0] >= 0 && i + dir[k][0] < M && j + dir[k][1] >= 0 && j + dir[k][1] < N)34                     gauss[i * N + j][(i + dir[k][0]) * N + j + dir[k][1]] = 1;35     int now = 0;36     for(int i = 0 ; i < M * N ; i++){37         int j = now;38         while(j < M * N && !gauss[j][i])39             j++;40         if(j == M * N)41             continue;42         if(j != now)43             swap(gauss[now] , gauss[j]);44         while(++j < M * N)45             if(gauss[j][i])46                 gauss[j] ^= gauss[now];47         now++;48     }49     for(int i = M * N - 1 ; i >= 0 ; i--){50         if(!gauss[i][i])51             ans[i] = 1;52         if(ans[i])53             for(int j = i - 1 ; j >= 0 ; j--)54                 if(gauss[j][i])55                     ans[j] = ans[j] ^ 1;56     }57     for(int i = 0 ; i < M ; i++){58         for(int j = 0 ; j < N ; j++){59             putchar(ans[i * N + j] + 48);60             putchar(‘ ‘);61         }62         putchar(‘\n‘);63     }64     return 0;65 }

Luogu1_4 cqoi2014 harmony matrix exclusive or Gaussian Element