bzoj 3534 重建

矩阵树定理中邻接矩阵可以理解为那条边存在的概率是1，那么这道题邻接矩阵只要是概率就行了。

然后再乱搞一下就行了

没有搞明白相对误差和绝对误差……省选的时候说过多少遍？

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>

#define md
#define ll long long
#define inf (int) 1e9
#define eps 1e-10
#define N 55
using namespace std;
double a[N][N];
double ans=1;
void outit()
{
	for (int i=1;i<=3;i++)
	{
		for (int j=1;j<=3;j++)
		  printf("%.2lf ",a[i][j]);
		printf("\n");
	}
}
void solve(int n)
{
	//outit();
	int f=1;
	for (int i=1;i<=n;i++)
	{
		int k=i; double mx=fabs(a[i][i]);
		for (int j=i+1;j<=n;j++) if (fabs(a[j][i])>mx) {mx=fabs(a[j][i]); k=j; }
		if (k!=i)
		{
			f=-f;
			for (int j=i;j<=n;j++) swap(a[i][j],a[k][j]);
		}
		for (int j=i+1;j<=n;j++)
		{
			double p=a[j][i]/a[i][i];
			for (int k=i;k<=n;k++) a[j][k]-=p*a[i][k];
		}
		if (fabs(a[i][i])<eps) { printf("0.000000\n"); exit(0);}
	}
	//outit();
	for (int i=1;i<=n;i++) ans=ans*a[i][i];
	ans=ans*f;
	printf("%.10lf\n",fabs(ans));
}
	
int main()
{
	//freopen("p7.in","r",stdin); freopen("p.out","w",stdout);
	int n;
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
	{
	  double sum=0;
	  for (int j=1;j<=n;j++)
	  {
	  	scanf("%lf",&a[i][j]);
	  	if (a[i][j]>1-eps) a[i][j]-=eps;
	  	if (i<j) ans=ans*(1-a[i][j]);
	  	if (i!=j)  
		{
			a[i][j]=-(a[i][j])/(1-a[i][j]);
	  		sum+=a[i][j];
	  	}
	  }
	  a[i][i]-=sum;
	}
	solve(n-1);
	return 0;
}
/*
3
0 0 1
0 0 0
1 0 0
*/