Blog of RuSun

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

P2151 [SDOI2009]HH去散步

P2151 [SDOI2009]HH去散步

边与点的转化思想,因为注意到不能相邻两次走一条边和其反向边,所以不能直接直接按照一般的点的做法。考虑如何避免从一条边走回反向边,考虑将其作为两条边,两条边没有路径即可。因此将所有的边作为点,重新建图即可。最后统计答案,只做 $t - 1$ 次乘法,因为从一条边的起点到一条边的终点也需要一个时间,对于两条边,如果一条边起点为起点,另一条边终点为终点,那么贡献答案。

查看代码
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#include <cstdio>
using namespace std;
const int N = 130, mod = 45989;
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');
}
struct Matrix
{
int n, m, w[N][N];
Matrix()
{
}
Matrix(int x, int y)
{
n = x, m = y;
}
Matrix(int x, int y, int k)
{
n = x, m = y;
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
w[i][j] = i == j ? k : 0;
}
void output ()
{
printf("%d %d\n", n, m);
for (int i = 0; i < n; i++, puts(""))
for (int j = 0; j < m; j++, putchar(' '))
write(w[i][j]);
}
friend Matrix operator*(Matrix x, Matrix y)
{
Matrix res(x.n, y.m, 0);
for (int i = 0; i < res.n; i++)
for (int j = 0; j < res.m; j++)
for (int k = 0; k < x.m; k++)
(res.w[i][j] += x.w[i][k] * y.w[k][j] % mod) %= mod;
return res;
}
friend Matrix operator^(Matrix b, int k)
{
Matrix res(b.n, b.m, 1);
while (k)
{
k & 1 && (res = res * b, 0);
b = b * b;
k >>= 1;
}
return res;
}
};
struct Edge
{
int u, v;
} e[N];
int main ()
{
int n, m, d, st, ed;
read(n), read(m), read(d), read(st), read(ed);
static Matrix mat(m << 1, m << 1);
for (int i = 0, a, b; i < m; i++)
{
read(a), read(b);
e[i << 1] = (Edge){a, b};
e[i << 1 | 1] = (Edge){b, a};
}
for (int i = 0; i < m << 1; i++)
for (int j = 0; j < m << 1; j++)
if (i ^ j ^ 1 && e[i].v == e[j].u)
mat.w[i][j] = 1;
mat = mat ^ d - 1;
int res = 0;
for (int i = 0; i < m << 1; i++)
for (int j = 0; j < m << 1; j++)
e[i].u == st && e[j].v == ed && ((res += mat.w[i][j]) %= mod);
write(res);
return 0;
}