int main ()
{
int n = 0;
initM ();
/** 一直训练直到能够100%正确为止 **/
while (1) {
n ++;
calcY ();
int err = adjustM ();
if (0 >= err) {
/** 能够 100 %正确地回答问题了，结束训练 **/
break;
}
printf ("错误数 %d\n", err);
}
printM ();
printf ("阈值: %d, 训练次数: %d\n", ST, n);
while (1) {
int a = 0;
scanf ("%i", &a);
if (0 > a || 9 < a) {
break;
}
test (a);
}
return 0;
}
void main() {
initM(0);
popularRedeSocialAleatoriamente(0.6);
imprimirRedeSocial();
int n = numAmigosEmComum(2,4);
printf("\nnumero de amigos em comum entre 2 e 4: %d", n);
//se voce eh forte, remova o comentario da linha abaixo
printf("\ncoef. de aglomeracao da pessoa 2 eh: %f", coeficienteAglomeracao(2));
getch();
}
double CDescriptor::ComputeLineDir(double* pLinePts,int nCount,double dDxAvg, double dDyAvg)
{
//利用最小二乘技术求方向
initM(MATCOM_VERSION);
Mm mMatrix = zeros(nCount,3);
for(int g1 = 0; g1 < nCount; g1++)
{
mMatrix.r(g1+1,1) = pLinePts[2*g1];
mMatrix.r(g1+1,2) = pLinePts[2*g1+1];
mMatrix.r(g1+1,3) = 1;
}
//奇异值分解获得精确位置
Mm u,s,v;
i_o_t i_o = {0,0};
svd(mMatrix,i_o,u,s,v);
//计算方向
double a = v.r(1,3);
double b = v.r(2,3);
double dMainArc = atan2(-b,a);
dMainArc = LimitArc(dMainArc);
//退出
exitM();
//判定方向
double dMainArc1 = dMainArc - PI/2;
dMainArc1 = LimitArc(dMainArc1);
double dMainArc2 = dMainArc + PI/2;
dMainArc2 = LimitArc(dMainArc2);
double dAvgArc = atan2(-dDyAvg,-dDxAvg);
dAvgArc = LimitArc(dAvgArc);
double error1 = ArcDis(dMainArc1,dAvgArc);
double error2 = ArcDis(dMainArc2,dAvgArc);
//返回最终方向
double nArcReturn = dMainArc1;
if(error1 > error2)
nArcReturn = dMainArc2;
return nArcReturn;
}
