The application of traceback in assigning the tasks

The best solutions:
  The 1th will arrange the tasks: 2
  The 2th will arrange the tasks: 1
  The 3th will arrange the tasks: 3
  The 4th will arrange the tasks: 4
   The total cost is: 13

The corresponding codes:
        这其实是书本上的代码，我主要对于第32行的代码感觉有点问题，书上对应的是：d[j] = 0;而我认为是的是d[i] =0; 我修改了一下，以后发现结果是一摸一样的，这也解答了我之前的一个疑惑，其实对于回溯而言，真正起到回到上一个结点的代码是递归函数的调用本身，而这个地方起到回到之前真正作用的是worker对应的取值，只要这个改变成之前原来的样子就ok了。

# include <stdio.h>
# include <string.h>
# include <queue>
# include <vector>
using namespace std;
# define INF 99999
# define MAXN 21
int n = 4;
int c[MAXN][MAXN] = {
   
   {0},{0,9,2,7,8},{0,6,4,3,7},{0,5,8,1,8},{0,7,6,9,4}};
int x[MAXN];
int cost = 0;
int bestx[MAXN];
int mincost = INF;
bool worker[MAXN];
void dfs(int i)
{
    if (i > n){
        if (cost < mincost){
            mincost = cost;
            for (int j = 1; j <= n; j++){
                bestx[j] = x[j];
            }
        }
    }else{
        for (int j = 1; j <= n; j++){
            if(!worker[j]){
                worker[j] = true;
                x[i] = j;
                cost += c[i][j];
                dfs(i+1);
                worker[j] = false;
                x[i] = 0;
                cost -= c[i][j];
            }
        }
    }
}
int main() {
    memset(worker, 0, sizeof(worker));
    dfs(1);
    printf("The best solutions:\n");
    for (int k =1 ; k <= n; k++) {
        printf("  The %dth will arrange the tasks: %d\n", k, bestx[k]) ;
    }
    printf("   The total cost is: %d\n", mincost);
}