[Poj 1308] Is it a tree? (Check the set to determine whether the graph is a tree with roots)

Description

A tree is a well-known data structure that is either empty (null, void, nothing) or is a set of one or more nodes connected by directed edges between nodes satisfying the following properties.

There is exactly one node, called the root, to which no directed edges point.
Every node should t the root has exactly one edge pointing to it.
There is a unique sequence of directed edges from the root to each node.
For example, consider the specified strations below, in which nodes are represented by circles and edges are represented by lines with arrowheads. The first two of these are trees, but the last is not.

In this problem you will be given several descriptions of collections of nodes connected by directed edges. For each of these you are to determine if the collection satisfies the definition of a tree or not.

Input

The input will consist of a sequence of descriptions (test cases) followed by a pair of negative integers. each test case will consist of a sequence of edge descriptions followed by a pair of zeroes each edge description will consist of a pair of integers; the first integer identifies the node from which the edge begins, and the second integer identifies the node to which the edge is directed. node numbers will always be greater than zero.

Output

For each test case display the line "case K is a tree. "or the line" case K is not a tree. ", where k corresponds to the test case number (they are sequentially numbered starting with 1 ).

Sample Input

6 8  5 3  5 2  6 45 6  0 08 1  7 3  6 2  8 9  7 57 4  7 8  7 6  0 03 8  6 8  6 45 3  5 6  5 2  0 0-1 -1

Sample output

Case 1 is a tree.Case 2 is a tree.Case 3 is not a tree.

Source

North central North America 1997

The shadow of the query set is completely invisible... This is to simulate a graph by checking the relationship between parent and child nodes. During the input process, if a child node is connected by multiple parent nodes, or the edge of a node points to itself, the root tree is not directly identified. If the root tree cannot be determined during the input process, after the parent-child relationship of the entire graph is established, you can also check the set to find the root node of each node to determine whether there is a root tree. If the root node of each node is different, it indicates that there are multiple root nodes, not a root tree, if the number of recursion times exceeds a certain size (1000) in the query process, it can be viewed as an endless loop, indicating that the graph has a ring and it can be determined that it is not a root tree. There are many restrictions on this question.

# Include <iostream> # define maxn 1010 using namespace STD; struct node {int F, used; // F = parent node, used = 1 indicates that the vertex has been connected to another vertex} node [maxn]; void makeset () {for (INT I = 0; I <maxn; I ++) {node [I]. f = 0; node [I]. used = 0 ;}} int findset (INT num, int step) {If (Step> 1000) Return-5; // If (node [num]. F = 0) return num; return findset (node [num]. f, step + 1);} void addedge (int A, int B) // create an directed edge a-> B {node [A]. used = 1; node [B]. used = 1; node [B]. F = A ;}int main () {int S, T, Case = 0; while (1) {bool istree = true; Case ++; makeset (); while (CIN> S> T & S> 0) {If (S = T) istree = false; else if (node [T]. F = 0) addedge (S, T); // The subnode t is not connected to any other vertex. You can add an else istree = false edge; // otherwise, the sub-node T is connected to another vertex, not a tree} If (S = 0) {If (! Istree) // It is determined that it is not a root tree {cout <"case" <Case <"is not a tree. "<Endl; continue;} int root = 0, Times = 0; For (INT I = 1; I <1000; I ++) {If (node [I]. used = 1) // point I and other edge connections {If (Times = 0) // first search for root = findset (I, 1 ); // determine a root node root else if (root! = Findset (I, 1) | findset (I, 1) <0 | root <0) // if the root node of vertex I is not a previously determined root node (with multiple root nodes) or has a ring, it is not a tree {istree = false; break; // No need to find} Times ++;} If (istree) cout <"case" <Case <"is a tree. "<Endl; else cout <" case "<Case <" is not a tree. "<Endl;} else break;} return 0 ;}

Zookeeper

[Poj 1308] Is it a tree? (Check the set to determine whether the graph is a tree with roots)