语音识别中的DTW算法问题

        之前处理语音信息的时候，在网上找过相关资料，但是质量良莠不齐。这里，把自己以前写过的DTW算法与大家分享。理论知识，大家可以参考：https://www.cnblogs.com/flypiggy/p/3603192.html，https://www.cnblogs.com/flypiggy/p/3603192.html

#ifndef DTW_H
#define DTW_H
#include <iostream>
#include <vector>
using namespace std;
//we get the minimum from three data
double MIN3(double fdata,double sdata,double tdata)
{
    double min1 = 0.0;
    //double min2 = 0.0;
    double result_min = 0.0;
    min1 = fdata > sdata ? sdata:fdata;
    result_min = tdata > min1 ? min1:tdata;
    return result_min;
}


//we can get the dtw between two speech segment;
double dtw(vector<double> feature_matr1,int fm1_row,int fm1_col,vector<vector<double> > &feature_matr2,int fm2_row,int fm2_col)
{
    int i,j,k,len;
    double dist,tsub,tsum;
    /*Initialize*/
    //double *tempdist = new double[fm1_row * fm2_row];
    vector<double> tempdist(fm1_row * fm2_row,0);
    //double *dp = new double[fm1_row * fm2_row];
    vector<double> dp(fm1_row * fm2_row,0);
    if(fm1_col != fm2_col)//第1、2个矩阵的列数不相等
    {
        return -1;
    }
    len = fm1_row * fm2_row;
    for (i = 0; i < fm1_row; i++)
        for (j = 0; j < fm2_row; j++)
        {   //下面计算的是第一个矩阵的第i行向量和第二个矩阵的j行向量的相似性，并储存在新建矩阵的[i][j]位置。
            tsum = 0.0;//compute the Euclidean distance between two vector of frames
            for (k = 0; k < fm2_col; k++)
            {
                tsub = feature_matr1[i*fm2_col+k] - feature_matr2[j][k];
                tsum+= tsub * tsub;
            }
            tempdist[i*fm2_row+j] = tsum;
        }

    /*DTW*/
    //We begin from top-left-corner, and the only direction are down,right,right-down.
    dp[0] = tempdist[0];//top-left-corner initialize
    for (i =1; i < fm1_row; i++)
        dp[i*fm2_row] = tempdist[i*fm2_row] + dp[(i-1)*fm2_row];
    for (i =1; i < fm2_row; i++)
        dp[i] = tempdist[i] + dp[i-1];
    for(i=1; i<fm1_row; i++)
    {
        for(j=1; j<fm2_row; j++)
        {
            dp[i*fm2_row+j] = tempdist[i*fm2_row+j] +
                                MIN3(dp[(i-1)*fm2_row+j],dp[(i-1)*fm2_row+j-1],dp[i*fm2_row+j-1]);
        }
    }
    dist = dp[(i-1)*fm2_row+j-1];
    return dist;
}

#endif // DTW_H