生成树计数

生成树计数，取摸//HDU4305#include <stdio.h>#include <algorithm>#include <iostream>#include <string.h>#include <vector>#include <queue>#include <map>#include <set>#include <list>#include <string>#include <math.h>using namespace std;struct Point{    int x,y;    Point(int _x = 0,int _y = 0)    {        x = _x,y = _y;    }    Point operator - (const Point &b)const    {        return Point(x-b.x,y-b.y);    }    int operator ^(const Point &b)const    {        return x*b.y - y*b.x;    }    void input()    {        scanf("%d%d",&x,&y);    }};struct Line{    Point s,e;    Line(){}    Line(Point _s,Point _e)    {        s = _s;        e = _e;    }};bool onSeg(Point P,Line L){    return    ((L.s-P)^(L.e-P)) == 0 &&    (P.x-L.s.x)*(P.x-L.e.x) <= 0 &&    (P.y-L.s.y)*(P.y-L.e.y) <= 0;}int sqdis(Point a,Point b){    return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);}const int MOD = 10007;int INV[MOD];//求ax = 1( mod m) 的x值，就是逆元(0<a<m)long long inv(long long a,long long m){    if(a == 1)return 1;    return inv(m%a,m)*(m-m/a)%m;}struct Matrix{    int mat[330][330];    void init()    {        memset(mat,0,sizeof(mat));    }    int det(int n)//求行列式的值模上MOD，需要使用逆元    {        for(int i = 0;i < n;i++)            for(int j = 0;j < n;j++)                mat[i][j] = (mat[i][j]%MOD+MOD)%MOD;        int res = 1;        for(int i = 0;i < n;i++)        {            for(int j = i;j < n;j++)                if(mat[j][i]!=0)                {                    for(int k = i;k < n;k++)                        swap(mat[i][k],mat[j][k]);                    if(i != j)                        res = (-res+MOD)%MOD;                    break;                }            if(mat[i][i] == 0)            {                res = -1;//不存在(也就是行列式值为0)                break;            }            for(int j = i+1;j < n;j++)            {                //int mut = (mat[j][i]*INV[mat[i][i]])%MOD;//打表逆元                int mut = (mat[j][i]*inv(mat[i][i],MOD))%MOD;                for(int k = i;k < n;k++)                    mat[j][k] = (mat[j][k]-(mat[i][k]*mut)%MOD+MOD)%MOD;            }            res = (res * mat[i][i])%MOD;        }        return res;    }};Point p[330];int n,R;bool check(int k1,int k2)//判断两点的距离小于等于R，而且中间没有点阻隔{    if(sqdis(p[k1],p[k2]) > R*R)return false;    for(int i = 0;i < n;i++)        if(i!=k1 && i!=k2)            if(onSeg(p[i],Line(p[k1],p[k2])))                return false;    return true;}int g[330][330];int main(){    //预处理逆元    for(int i = 1;i < MOD;i++)        INV[i] = inv(i,MOD);    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&R);        for(int i = 0;i < n;i++)            p[i].input();        memset(g,0,sizeof(g));        for(int i = 0;i < n;i++)            for(int j = i+1;j <n;j++)                if(check(i,j))                    g[i][j] = g[j][i] = 1;        Matrix ret;        ret.init();        for(int i = 0;i < n;i++)            for(int j = 0;j < n;j++)                if(i != j && g[i][j])                {                    ret.mat[i][j] = -1;                    ret.mat[i][i]++;                }        printf("%d\n",ret.det(n-1));    }    return 0;}

生成树计数 不取摸//spoj104来自 http://blog.csdn.net/u011699990/article/details/45182125#include <cstdio>#include <iostream>#include <cstring>#include <cmath>#include <algorithm>#include <string.h>#include <string>#include <vector>#include <queue>#define MEM(a,x) memset(a,x,sizeof a)#define eps 1e-8#define MOD 10009#define MAXN 110#define MAXM 100010#define INF 99999999#define ll __int64#define bug cout<<"here"<<endl#define fread freopen("ceshi.txt","r",stdin)#define fwrite freopen("out.txt","w",stdout)using namespace std;int sgn(double x){    if(fabs(x)<eps) return 0;    if(x<0) return -1;    else return 1;}double b[MAXN][MAXN];double det(double a[][MAXN],int n){    int i,j,k,sign=0;    double ret=1;    for(i=0;i<n;i++)        for(j=0;j<n;j++)            b[i][j]=a[i][j];    for(i=0;i<n;i++){        if(sgn(b[i][i])==0){            for(j=i+1;j<n;j++)                if(sgn(b[j][i])!=0)                    break;            if(j==n) return 0;            for(k=i;k<n;k++)                swap(b[i][k],b[j][k]);            sign++;        }        ret*=b[i][i];        for(k=i+1;k<n;k++)            b[i][k]/=b[i][i];        for(j=i+1;j<n;j++)            for(k=i+1;k<n;k++)                b[j][k]-=b[j][i]*b[i][k];    }    if(sign&1) ret=-ret;    return ret;}double a[MAXN][MAXN];int g[MAXN][MAXN];int main(){    int tc;    scanf("%d",&tc);    while(tc--){        int n,m;        scanf("%d%d",&n,&m);        MEM(g,0);        while(m--){            int u,v;            scanf("%d%d",&u,&v);            u--; v--;            g[u][v]=g[v][u]=1;        }        MEM(a,0);        for(int i=0;i<n;i++)            for(int j=0;j<n;j++)                if(i!=j&&g[i][j]){                    a[i][i]++;                    a[i][j]=-1;                }        double ans=det(a,n-1);        printf("%.0lf\n",ans);    }    return 0;}