使用动态编程的tsp

我当时使用一段代码来使用动态编程来实现TSP。我已经找到了此代码，但无法弄出compute（）函数及其工作方式。我不知道什么是变量，以及它如何计算路径。任何帮助都将受到高度赞赏。

#include <stdio.h>
#include<limits.h>
#define size 10 //maximum 10 cities
#define min(a,b) a>b?b:a
#define sizePOW 1024
int n,npow,g[size][sizePOW],p[size][sizePOW],adj[size][size];
int compute(int start,int set)
{   
int masked,mask,result=INT_MAX,temp,i,t1;//result stores the minimum 
if(g[start][set]!=-1)//memoization DP top-down,check for repeated subproblem
    return g[start][set];
for(i=0;i<n;i++)
    {   //npow-1 because we always exclude "home" vertex from our set

        t1=1<<i;
        mask=(npow-1)-(1<<i);//remove ith vertex from this set
        masked=set&mask;
        if(masked!=set)//in case same set is generated(because ith vertex was not present in the    set hence we get the same set on removal) eg 12&13=12
        {   
            temp=adj[start][i]+compute(i,masked);//compute the removed set
            if(temp<result)
                result=temp,p[start][set]=i;//removing ith vertex gave us minimum
        }
    }
    return g[start][set]=result;//return minimum
}

void TSP()
{     
int i,j;
//g(i,S) is length of shortest path starting at i visiting all vertices in S and ending at 1
for(i=0;i<n;i++)
    for(j=0;j<npow;j++) 
            g[i][j]=p[i][j]=-1; 
for(i=0;i<n;i++){
    g[i][0]=adj[i][0];//g(i,nullset)= direct edge between (i,1)
}

int result=compute(0,npow-2);//npow-2 to exclude our "home" vertex
printf("Tour cost:%d\n",result);
printf("Tour path:\n0 ");
getpath(0,npow-2);
printf("0\n");
}
int main(void) {
int i,j;
printf("Enter number of cities\n");
scanf("%d",&n);
npow=(int)pow(2,n);//bit number required to represent all possible sets
printf("npow = %d   ",npow);
printf("\nEnter the adjacency matrix\n");

for(i=0;i<n;i++)for(j=0;j<n;j++){
scanf("%d",&adj[i][j]);}
TSP();

return 0;
}