[HEOI2015]小Z的房间 && [CQOI2018]社交网络

今天看了一下矩阵树定理，然后学了一下\(O(n ^ 3)\)的方法求行列式。
哦对了，所有的证明我都没看……


这位大佬讲的好呀：
[学习笔记]高斯消元、行列式、Matrix-Tree 矩阵树定理


关于模数不是质数的情况，我看了半天才懂：其实就是加速了两行的辗转相减，把一次次减换成了取模。然后别忘了每一次交换两行的时候行列式要取相反数。


[HEOI2015]小Z的房间
这题就是板儿题了。把是'.'的格子记录下来，然后构造基尔霍夫矩阵就行了。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 105;
const ll mod = 1e9;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

int n, m, cnt = 0, id[maxn][maxn];
char s[maxn][maxn];
ll f[maxn][maxn];

In void add(int x, int y)
{
  ++f[x][x], ++f[y][y], --f[x][y], --f[y][x];
}

In ll Gauss(int n)
{
  ll ans = 1;
  for(int i = 1; i <= n; ++i)
    {
      for(int j = i + 1; j <= n; ++j)
    while(f[j][i])
      {
        ll d = f[i][i] / f[j][i];
        for(int k = i; k <= n; ++k) f[i][k] = (f[i][k] - d * f[j][k] % mod + mod) % mod;
        swap(f[i], f[j]), ans = -ans;
      }
      ans = (ans * f[i][i] % mod + mod) % mod;
    }
  return ans;
}

int main()
{
  n = read(), m = read();
  for(int i = 1; i <= n; ++i) scanf("%s", s[i] + 1);
  for(int i = 1; i <= n; ++i)
    for(int j = 1; j <= m; ++j) if(s[i][j] == '.') id[i][j] = ++cnt;
  for(int i = 1; i <= n; ++i)
    for(int j = 1; j <= m; ++j)
      if(id[i][j])
    {
      if(id[i][j + 1]) add(id[i][j], id[i][j + 1]);
      if(id[i + 1][j]) add(id[i][j], id[i + 1][j]);
    }
  write(Gauss(cnt - 1)), enter;
  return 0;
}

[CQOI2018]社交网络
这题让求的是以节点1为根的生成树个数，于是我们把矩阵的第一行第一列消去，剩下的求一个行列式就是答案了。
注意连边是反的。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 255;
const ll mod = 1e4 + 7;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

int n, m;
ll f[maxn][maxn];

In ll quickpow(ll a, ll b)
{
  ll ret = 1;
  for(; b; b >>= 1, a = a * a % mod)
    if(b & 1) ret = ret * a % mod;
  return ret;
}

In ll Gauss()
{
  ll ans = 1;
  for(int i = 2; i <= n; ++i)
    {
      int pos = i;
      while(!f[pos][i]) ++pos;
      if(pos ^ i) swap(f[i], f[pos]), ans = mod - ans;
      if(!f[i][i]) return 0;  //如果每一行第i个数都输0,才返回0
      ans = ans * f[i][i] % mod;
      ll inv = quickpow(f[i][i], mod - 2);
      for(int j = i + 1; j <= n; ++j)
    {
      ll tp = f[j][i] * inv % mod;
      for(int k = i; k <= n; ++k) f[j][k] = (f[j][k] - tp * f[i][k] % mod + mod) % mod;
    }
    }
  return ans;
}

int main()
{
  n = read(), m = read();
  for(int i = 1; i <= m; ++i)
    {
      int y = read(), x = read();  //x -> y
      ++f[y][y], --f[y][x];
    }
  write(Gauss()), enter;
  return 0;
}