Blog of RuSun

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

P6442 [COCI2011-2012#6] KOŠARE

P6442 [COCI2011-2012#6] KOŠARE

与 CF449D Jzzhu and Numbers 很相似。

查看代码
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#include <cstdio>
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 = 20, M = 1 << N, mod = 1e9 + 7;
int n, m, p[M], f[M];
int main ()
{
read(n), read(m);
p[0] = 1;
for (int i = 1; i <= n; i++)
p[i] = (p[i - 1] << 1) % mod;
for (int k; n; n--)
{
read(k);
int s = 0;
for (int a; k; k--)
{
read(a);
s |= 1 << a - 1;
}
f[s]++;
}
for (int i = 0; i < m; i++)
for (int j = 0; j < 1 << m; j++)
j >> i & 1 && (f[j] += f[j ^ 1 << i]);
for (int i = 0; i < 1 << m; i++)
f[i] = p[f[i]] - 1;
for (int i = 0; i < m; i++)
for (int j = 0; j < 1 << m; j++)
j >> i & 1 && ((f[j] -= f[j ^ 1 << i]) %= mod);
write((f[(1 << m) - 1] + mod) % mod);
return 0;
}