Blog of RuSun

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

CF963E Circles of Waiting

LuoGu: CF963E Circles of Waiting

CF: E. Circles of Waiting

高斯消元,直接上的话复杂度 $O(r ^ 6)$ ,但是发现范围很集中,用带状矩阵高斯消元只用消 $r$ 内的,复杂度 $O(r ^ 4)$ ,空间也很卡。

查看代码
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#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long LL;
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 = 7846, M = 110, mod = 1e9 + 7, dx[] = {-1, 0, 1, 0}, dy[] = {0, -1, 0, 1};
int n, id[M][M], A[N][N];
int binpow (int b, int k = mod - 2)
{
int res = 1;
while (k)
{
k & 1 && (res = (LL)res * b % mod);
b = (LL)b * b % mod;
k >>= 1;
}
return res;
}
void Gauss (int d)
{
for (int k = 0; k < n; k++)
{
int t = min(n - 1, k + d), inv = binpow(A[k][k]);
A[k][n] = (LL)A[k][n] * inv % mod;
for (int i = t; i >= k; i--)
A[k][i] = (LL)A[k][i] * inv % mod;
for (int i = k + 1; i <= t; i++)
{
(A[i][n] -= (LL)A[k][n] * A[i][k] % mod) %= mod;
for (int j = t; j >= k; j--)
(A[i][j] -= (LL)A[k][j] * A[i][k] % mod) %= mod;
}
}
for (int i = n - 1; ~i; i--)
for (int j = i + 1; j < n; j++)
(A[i][n] -= (LL)A[i][j] * A[j][n] % mod) %= mod;
}
int main ()
{
int r, p[4], inv = 0;
read(r);
for (int i = 0; i < 4; i++)
read(p[i]), (inv += p[i]) %= mod;
inv = binpow(inv);
for (int i = 0; i < 4; i++)
p[i] = (LL)p[i] * inv % mod;
for (int i = 0; i <= r << 1; i++)
for (int j = 0; j <= r << 1; j++)
id[i][j] = (i - r) * (i - r) + (j - r) * (j - r) > r * r ? -1 : n++;
int d = 0;
for (int i = 0; i <= r << 1; i++)
for (int j = 0; j <= r << 1; j++)
{
if (id[i][j] == -1)
continue;
for (int k = 0; k < 4; k++)
{
int nx = i + dx[k], ny = j + dy[k];
if (nx < 0 || ny < 0 || nx > r << 1 || ny > r << 1 || id[nx][ny] == -1)
continue;
A[id[nx][ny]][id[i][j]] = -p[k];
d = max(d, abs(id[i][j] - abs(id[nx][ny])));
}
A[id[i][j]][id[i][j]] = 1, A[id[i][j]][n] = 1;
}
Gauss(d);
write((A[id[r][r]][n] + mod) % mod);
return 0;
}

更优秀的做法是主元法,咕咕咕。