Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P3164 [CQOI2014]和谐矩阵

P3164 [CQOI2014]和谐矩阵

直接上高斯消元。因为一定有多个解,所以一定有自由元,因为不允许全部为 $0$ ,所以令自由元答案为 $1$ 即可,并参与后面的消元。

查看代码
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#include <cstdio>
#include <bitset>
using namespace std;
template <class Type>
void read(Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 3) + (x << 1) + c - '0';
flag && (x = ~x + 1);
}
template <class Type>
void write(Type x)
{
x < 0 && (putchar('-'), x = ~x + 1);
x > 9 && (write(x / 10), 0);
putchar(x % 10 + '0');
}
const int N = 1610, dx[] = {0, 1, -1, 0}, dy[] = {1, 0, 0, -1};
int n, m, tot;
bitset<N> A[N];
void Gauss ()
{
for (int k = 1; k <= tot; k++)
{
int t = k;
while (t <= tot && !A[t][k])
t++;
t > tot ? A[k][0] = 1 : (swap(A[t], A[k]), 0);
for (int i = 1; i <= tot; i++)
i ^ k && A[i][k] && (A[i] ^= A[k], 0);
}
}
int num (int a, int b)
{
return (a - 1) * m + b;
}
bool inside (int a, int b)
{
return a > 0 && a <= n && b > 0 && b <= m;
}
int main ()
{
read(n), read(m);
tot = n * m;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
{
int p = num(i, j);
for (int k = 0; k < 4; k++)
{
int nx = i + dx[k], ny = j + dy[k];
if (inside(nx, ny))
A[p][num(nx, ny)] = 1;
}
A[p][p] = 1;
}
Gauss();
for (int i = 1; i <= n; i++, puts(""))
for (int j = 1; j <= m; j++, putchar(' '))
write(A[num(i, j)][0]);
return 0;
}