如何实现对人脸图像数据库中的某一图像进行查找、对比、分析？

Question

例如有一个人脸图像库，如何利用摄像头即时捕捉的人脸图像，
与已经保存的人脸数据库中查找、对比、辨别等，希望能得到的实现思想即可。  希望与对此感兴趣的网友交流！

Answer 1

俺前天才完成的3层BP网模板,就为你的10分拿出来抛砖引玉.const int N0=32,N1=14,N2=4;    //------结点数
double a1[(N1+1)*(N0+1)],a2[N2*(N1+1)];//-------权值
double x[(N0+1)];//------输入
double a;//----logistic 函数参数
double o1[(N1+1)],o2[N2];//-------各神经元输出标量
double s1[(N1+1)],s2[N2];//---------局部梯度域
double n;     //-----------学习率
//----------网络初始化
        //---------初始化特征
        for(int i=0;i<(N1+1)*(N0+1);i++)
        {
                a1[i]=1;
        }
        for(int i=0;i<(N2)*(N1+1);i++)
        {
                a2[i]=1;
        }
        a=1;//----logistic 函数参数
        n=0.8;  //---------学习率//----------------神经网络
//---------计算隐藏层的输出/输出层端的输入
                        double v=0;
                        for(i=0;i<N1;i++)
                        {
                                v=0;
                                for(j=0;j<N0+1;j++)
                                {
                                        //线性求和
                                        v+=(a1[i*(N0+1)+j]*x[j]);
                                }
                                //logistic函数(sigmoid)
                                o1[i]=(double)1.0/(1.0+exp(-a*v));
                        }
                        o1[N1]=1;         //--------输出层偏置输入=1
                        //------------计算输出层输出
                        for(i=0;i<N2;i++)
                        {
                                v=0;
                                for(j=0;j<N1+1;j++)
                                {
                                        //线性求和
                                        v+=(a2[i*(N1+1)+j]*o1[j]);
                                }
                                //logistic函数(sigmoid)
                                o2[i]=(double)1.0/(1.0+exp(-a*v));
                        }
//----------------------------------------------
if(mode==0)
{
xp=5000;
                                int xx[N2];
                                //------对输出层神经元的输出量化到0或1
                                for(i=0;i<N2;i++)
                                {
                                        if(o2[i]<0.5)
                                                xx[i]=0;
                                        else
                                                xx[i]=1;
                                }
                                //----------2进制转10进制
                                int result=0;
                                for(i=0;i<N2;i++)
                                {
                                        result+=(xx[i]<<i);
                                }                                if(result<0)
                                        result=0;
                                if(result>9)
                                        result=9; CString s;
s.Format("%d",result);
red.Append(s);

}
else
{

char c[2];c[0]=cred.GetAt(count);c[1]=0;
int cre=atoi(c);
                                int d[4];
                                d[0]=cre%2;cre/=2;
                                d[1]=cre%2;cre/=2;
                                d[2]=cre%2;cre/=2;
                                d[3]=cre%2;cre/=2;                                //-------------输出层调节
                                //----------计算输出层的局部梯度域
                                for(i=0;i<N2;i++)
                                {
                                        s2[i]=a*((double)d[i]-o2[i])*o2[i]*(1-o2[i]);
                                }
                                ///-------------调整权值
                                for(i=0;i<N2;i++)
                                {
                                        for(j=0;j<N1+1;j++)
                                        {
                                                a2[i*(N1+1)+j]+=(n*s2[i]*o1[j]);
                                        }
                                }                                //----------------隐藏层调节
                                //----------计算隐藏层结点局部梯度域
                                for(i=0;i<N1+1;i++)
                                {
                                        double ek=0;
                                        for(j=0;j<N2;j++)
                                        {
                                                ek+=(s2[j]*a1[j*(N1+1)+i]);
                                        }
                                        s1[i]=a*o1[i]*(1-o1[i])*ek;
                                }
                                ///-------------调整权值
                                for(i=0;i<N1+1;i++)
                                {
                                        for(j=0;j<N0+1;j++)
                                        {
                                                a1[i*(N0+1)+j]+=(n*s1[i]*x[j]);
                                        }
                                }
                                count++;
}

Answer 2

神经网络的代码简单，关键是怎么训练，网络结构，另外，这个问题我觉得还是pca是王道

Answer 3

感谢sboom兄，分数虽少，望笑纳！