SOR法

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int SOR(vector< vector<double> > &A, int n, double w, int &max_iter, double &eps)
{
    vector<double> x(n, 0.0);        double tmp;
 
    double e = 0.0;
    int k;
    for(k = 0; k < max_iter; ++k){
                e = 0.0;
        for(int i = 0; i < n; ++i){
            tmp = x[i];
            x[i] = A[i][n];
            for(int j = 0; j < n; ++j){
                x[i] -= (j != i ? A[i][j]*x[j] : 0.0);
            }
            x[i] /= A[i][i];
 
            x[i] = tmp+w*(x[i]-tmp);
 
            e += fabs(tmp-x[i]);                        }
 
                if(e <= eps){
            break;
        }
    }
 
    max_iter = k;
    eps = e;
 
    for(int i = 0; i < n; ++i){
        A[i][n] = x[i];
    }
 
    return 1;
}