题目链接
题解
没想到……直接用暴力的(O((nm)^3))算法，居然能过？！
高斯消元解异或方程组。
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#define space putchar(' ')
#define enter putchar('
')
typedef long long ll;
using namespace std;
template <class T>
void read(T &x){
char c;
bool op = 0;
while(c = getchar(), c < '0' || c > '9')
if(c == '-') op = 1;
x = c - '0';
while(c = getchar(), c >= '0' && c <= '9')
x = x * 10 + c - '0';
if(op) x = -x;
}
template <class T>
void write(T x){
if(x < 0) putchar('-'), x = -x;
if(x >= 10) write(x / 10);
putchar('0' + x % 10);
}
const int N = 1605;
int n, m, x, g[N][N], ans[N];
const int dx[5] = {0, -1, 1, 0, 0};
const int dy[5] = {0, 0, 0, -1, 1};
void gauss(){
for(int i = 1; i <= x; i++){
if(!g[i][i])
for(int j = i + 1; j <= x; j++)
if(g[j][i]){
for(int k = 1; k <= x + 1; k++)
swap(g[i][k], g[j][k]);
break;
}
for(int j = i + 1; j <= x; j++)
if(g[j][i])
for(int k = i; k <= x + 1; k++)
g[j][k] ^= g[i][k];
}
for(int i = x; i; i--){
if(!g[i][i]) ans[i] = 1;
else{
for(int j = i + 1; j <= x; j++)
g[i][x + 1] ^= ans[j] & g[i][j];
ans[i] = g[i][x + 1];
}
}
}
int main(){
read(n), read(m);
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++){
x++;
for(int d = 0; d <= 4; d++){
int ti = i + dx[d], tj = j + dy[d];
if(ti <= n && tj <= m && ti && tj)
g[x][(ti - 1) * m + tj] = 1;
}
}
gauss();
for(int i = 1; i <= x; i++)
write(ans[i]), i % m ? space: enter;
return 0;
}