Please anybody help ..!!

Like the tags of the problem indicate, it's solved by using bitmasks and DP.

• The state of the DP is the objects put in the handbag so far. You represent these using bitmasks (objects have to be 0-indexed). For example, if you've put objects 0, 3 and 5, the state is 41 (2⁰ + 2³ + 2⁵). The size of the DP array will then be 2²⁴.
• Iterate through all the states from 0 to 2^N - 1 seeing if adding one or two objects to this state yields a better solution than what you currently have. The cost of adding two objects i and j (potentially the same) is Dist[H][i] + Dist[i][j] + Dist[j][H], where H is the handbag. Dist[i][j] = (x[i] - x[j])² + (y[i] - y[j])². Cost[i][i] = 0, obviously.
• One key observation to make it run in time is that the order in which you put the objects doesn't matter. For example, if the optimal solution is 0 4 3 0 1 0 2 0, it will be the same as 0 1 0 3 4 0 2 0 and the same as 0 1 0 2 0 3 4 0. So, for every state, only consider adding the first object that hasn't been added yet (and maybe another one together with it).
• To reconstruct the optimal "path", when you find that you can reach state 'y' from state 'x' with optimal cost, store that in an auxiliary array (i.e., From[y] = x).