【SPOJ104】Highways

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题目大意：

有N个节点，某些节点之间可以连边，求有多少种方案可以使任意两点间有唯一路径

【分析】

生成树计数问题：Marix-Tree定理

具体含义：http://download.csdn.net/detail/u012274244/6638281

【代码】

/*    ID:CiocioLANG:C++DATE:2013-11-30TASK:SPOJ-104 Highways*/#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <iostream>#include <algorithm>#include <queue>using namespace std;#define MAXN 20#define LL long longint N,M,degree[MAXN];LL C[MAXN][MAXN];void _init(){scanf("%d%d",&N,&M);for(int i=1;i<=M;i++){int x,y;scanf("%d%d",&x,&y);C[x][y]=C[y][x]=-1;degree[x]++;degree[y]++;}}LL _det(LL a[MAXN][MAXN],int n){LL ret=1;for(int i=1;i<=n;i++){for(int j=i+1;j<=n;j++)while(a[j][i]){LL t=a[i][i]/a[j][i];for(int k=i;k<=n;k++){a[i][k]=(a[i][k]-a[j][k]*t);swap(a[i][k],a[j][k]);}ret*=(-1);}if(!a[i][i]) return 0;ret*=a[i][i];}return abs(ret);}void _solve(){for(int i=1;i<=N;i++)     //C即为原图的Kirchhoff矩阵C[i][i]=degree[i];printf("%lld\n",_det(C,N-1));   //对矩阵C求一次N-1阶主子式的行列式绝对值}int main(){int Case;scanf("%d",&Case);while(Case--){memset(C,0,sizeof(C));memset(degree,0,sizeof(degree));    _init();    _solve();}return 0;}