spoj104 highways（矩阵树定理）

最新推荐文章于 2018-08-04 08:29:43 发布

yukang_0613

最新推荐文章于 2018-08-04 08:29:43 发布

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分类专栏：矩阵树定理

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本文介绍了一种利用矩阵树定理解决特定高速公路网络构建问题的方法，该问题要求计算连接多个城市的高速公路网络中，确保任意两城市间存在唯一路径的不同方案数量。

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HIGH - Highways

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In some countries building highways takes a lot of time... Maybe that's because there are many possiblities to construct a network of highways and engineers can't make up their minds which one to choose. Suppose we have a list of cities that can be connected directly. Your task is to count how many ways there are to build such a network that between every two cities there exists exactly one path. Two networks differ if there are two cities that are connected directly in the first case and aren't in the second case. At most one highway connects two cities. No highway connects a city to itself. Highways are two-way.

Input

The input begins with the integer t, the number of test cases (equal to about 1000). Then t test cases follow. The first line of each test case contains two integers, the number of cities (1<=n<=12) and the number of direct connections between them. Each next line contains two integers a and b, which are numbers of cities that can be connected. Cities are numbered from 1 to n. Consecutive test cases are separated with one blank line.

Output

The number of ways to build the network, for every test case in a separate line. Assume that when there is only one city, the answer should be 1. The answer will fit in a signed 64-bit integer.

Example

Sample input:
4
4 5
3 4
4 2
2 3
1 2
1 3

2 1
2 1

1 0

3 3
1 2
2 3
3 1

Sample output:
8
1
1
3

题目大意：给你一个点数为n，边数为m的无向图，问你如何使每两个点间有且只有一条路线有多少方案

可以很容易想到最小生成树，但是方案数又很难处理

通过看题解发现这道题可以用矩阵树定理（matrix tree）很轻松地解出来

矩阵树定理：

构造基尔霍夫矩阵，并设为C，使C=(D-G)（D：度数矩阵，G：联通矩阵）

把C变成C的任意（n-1）阶主子式，C的行列式就是答案

行列式求法（用高斯消元把矩阵化成上三角矩阵，然后把C[i][i]（1<=i<=n）乘起来就是答案

具体见代码

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
const int maxn=107;
double c[maxn][maxn];
int a[maxn][maxn],b[maxn][maxn];
int n;
double eps=1e-8;
double det(){
	int i,j,k,r;
	for(i=1;i<=n;i++)
	{
		r=i;
		for(j=i+1;j<=n;j++)
		if(fabs(c[j][i])>fabs(c[r][i])) r=j;
		if(c[r][i]<eps){
			return 0;
		}
		if(r!=i) for(j=1;j<=n;j++) swap(c[i][j],c[r][j]);
		for(j=i+1;j<=n;j++)
		{
			double f=c[j][i]/c[i][i];
			for(k=i;k<=n;k++){
				c[j][k]-=f*c[i][k];
			}
		}
	}
	double ans=1.0;
	if(n&1) ans=ans;
	for(int i=1;i<=n;i++)
	{
		ans*=c[i][i];
	}
	return ans;
}
int main(){
	int t;
	scanf("%d",&t);
	while(t--){		
		int m;
		scanf("%d%d",&n,&m);
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		n--;
		for(int i=1;i<=m;i++)
		{
			int u,v;
			scanf("%d%d",&u,&v);
			u--;
			v--;
			a[u][u]++;a[v][v]++;
			b[u][v]++;b[v][u]++;
		}
		for(int i=1;i<=n;i++)
		for(int k=1;k<=n;k++)
		c[i][k]=(double)a[i][k]-b[i][k];
		printf("%.0lf",det());
	}
	return 0;
}