travelling-salesman-problem

//Assuming that there are 20 friends
int dp[1<<20][20];
//mask=frinds I already visited
//at=last visited friend
int Dp(int mask, int at)
{
	//I used reference. It helps me not
	//writing dp[mask] [at] again and again.
	int& ret = dp[mask] [at];
	//Assume that we initilized dp with -1
	if (ret !=-1) return ret;

	//initialize ret with infinity
	ret = 1000000000;
	//n = number of friend
	//dist contains distance between every two nodes
	for (int i = 0; i<n; i++)
		inf (!(mask & (1<<i)))
		//mask | (1<<i) turns on i th bit in mask.
		ret = MIN(re,
			DP(mask | (1<<i), i) + dist[at][i]);
	return ret;
}