P4111 [HEOI2015] 小 Z 的房间 题解

这个大房子可以看成一张 $n\times m$ 的网格图，

柱子不能打通的限制可以看成删去每个柱子对应的点，

此时答案即为剩下的图的生成树个数。

用矩阵树定理求解即可。

#include <cstdio>
#include <algorithm>
#define M 1000000000
#define int long long
using namespace std;
char z;
int u, v, n, _ = 1, q = 1, o[350][350], a[350][350];
void A(int u, int v)
{
    a[u][u] = (a[u][u] + 1) % M, a[v][v] = (a[v][v] + 1) % M;
    a[u][v] = (a[u][v] + M - 1) % M, a[v][u] = (a[v][u] + M - 1) % M;
}
signed main()
{
    scanf("%lld%lld", &u, &v);
    for (int i = 1; i <= u; ++i)
        for (int j = 1; j <= v; ++j)
        {
            scanf(" %c", &z);
            if (z == '.')
                o[i][j] = ++n;
        }
    for (int i = 1; i <= u; ++i)
        for (int j = 1; j <= v; ++j)
        {
            if (i != u && o[i][j] && o[i + 1][j])
                A(o[i][j], o[i + 1][j]);
            if (j != v && o[i][j] && o[i][j + 1])
                A(o[i][j], o[i][j + 1]);
        }
    for (int i = 2; i <= n; ++i)
        for (int j = i + 1; j <= n; ++j)
        {
            while (a[i][i])
            {
                int z = a[j][i] / a[i][i];
                for (int k = i; k <= n; ++k)
                    a[j][k] = (a[j][k] + M - a[i][k] * z % M) % M;
                swap(a[i], a[j]);
                _ = -_;
            }
            swap(a[i], a[j]);
            _ = -_;
        }
    for (int i = 2; i <= n; ++i)
        q = (q * a[i][i]) % M;
    if (_ == -1)
        q = (M - q) % M;
    printf("%lld", q);
    return 0;
}