Question

我正在使用一段代码来使用动态编程来实现TSP。我找到了这段代码，但无法弄清楚compute（）函数及其工作原理。我不知道变量是什么以及它如何计算路径。任何帮助都非常感谢。

#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;
}

Answer 1

//g(i,S) is length of shortest path starting at i visiting all vertices in S and ending at 1

此评论非常重要。

图片目前我们在＆＃34;开始＆＃34;，我们访问的城市是＆＃34;设置＆＃34;（以二进制表示），我们如何计算其他状态？

例如，我们有一个优势＆＃34;开始＆＃34; - ＆gt;＆＃34; i＆＃34;，然后

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
    }

这是使用g [start] [set]和edge（start，i）计算状态g [start] [masked]

正如我们所看到的，作者使用递归和动态编程来解决问题。时间是O（2 ^ n * n ^ 2）

tsp使用动态编程

1 个答案: