Problem Description<\/h2>\n
One day, a special event for gift delivery was held in a village. The purpose of this event is to deliver gifts among different people. There are n<\/strong> participants, and each participant must deliver exactly one gift. However, there are constraints on whom they can deliver gifts to.<\/p>\n Each participant has a list or array that indicates who they can deliver their gifts to. Considering these constraints, we need to determine whether all participants can deliver gifts, meaning whether every participant can receive a gift.<\/p>\n The first line contains the number of participants n<\/strong>. The next line contains a list that represents the possible recipients for each of the n<\/strong> participants. Each element of the list expresses the index of the participant that the corresponding participant can deliver their gift to.<\/p>\n If all participants can receive gifts, print ‘YES’; otherwise, print ‘NO’.<\/p>\n This problem can be viewed similarly to the “cycle detection” problem in graph theory. Each participant can be represented as a node, and the relationships of whom they can give gifts to can be expressed as edges, forming a directed graph. All nodes must be visited at least once, and if the graph has only one cycle, we can return ‘YES’.<\/p>\n Based on the provided information, we construct the graph in list form. Each participant’s designated recipient is represented by their index in the list.<\/p>\n We can traverse the graph using DFS (Depth First Search) or BFS (Breadth First Search). The goal is to determine whether all participants can receive a gift. This involves checking each node for visit status while traversing the graph.<\/p>\n We check if a cycle exists that allows every participant to give and receive gifts. This cycle must include all nodes, and if there is exactly one cycle, everyone will be able to receive a gift.<\/p>\n Based on the above methods, we implement a Python code:<\/p>\n<\/section>\n The code above is an algorithm to verify gift delivery. The 'can_give_gifts' function takes the number of participants n<\/strong> and the list of possible gift receivers connections<\/strong> as arguments. A visited<\/strong> list is set up to check each participant's visit status. DFS is used to visit each participant, increasing the count. If all participants are visited, it returns 'YES'; otherwise, it returns 'NO'.<\/p>\n Let's examine this algorithm in action through the following example:<\/p>\n This problem is a common type encountered in Python coding tests. It helps in understanding graph concepts and the method of solving problems through cycle detection. By practicing various types of problems, you can enhance your insights into algorithms.<\/p>\n<\/article>\n <\/body><\/p>\n","protected":false},"excerpt":{"rendered":"Input Format<\/h2>\n
Output Format<\/h2>\n
Example Input<\/h2>\n
\n5\n1 2 3 4 0\n <\/pre>\n
Example Output<\/h2>\n
\nYES\n <\/pre>\n
Solution Method<\/h2>\n
Step-by-Step Solution Process<\/h2>\n
1. Constructing the Graph<\/h3>\n
2. Using DFS or BFS<\/h3>\n
3. Cycle Detection<\/h3>\n
4. Implementation<\/h3>\n
\ndef can_give_gifts(n, connections):\n visited = [False] * n\n count = 0\n\n def dfs(v):\n nonlocal count\n visited[v] = True\n count += 1\n next_giver = connections[v]\n if not visited[next_giver]:\n dfs(next_giver)\n\n for i in range(n):\n if not visited[i]:\n count = 0\n dfs(i)\n if count < n:\n return \"NO\"\n\n return \"YES\"\n\n# Test\nn = 5\nconnections = [1, 2, 3, 4, 0]\nprint(can_give_gifts(n, connections))\n <\/pre>\n
Code Explanation<\/h2>\n
Comprehensive Example<\/h2>\n
\n6\n1 2 3 4 5 0\n <\/pre>\n
Expected Output<\/h2>\n
\nYES\n <\/pre>\n
Conclusion<\/h2>\n