POJ 1308 Is It A Tree?

Is It A Tree?

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 18778		Accepted: 6395

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 except 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 illustrations 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 4
5 6  0 0

8 1  7 3  6 2  8 9  7 5
7 4  7 8  7 6  0 0

3 8  6 8  6 4
5 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.
題目大意：給出一些數字對，表明父節點和子節點，輸入以0 0結束，問這些節點可否構成一棵樹。
解題方法：判斷可否構成一棵樹的條件：1.空樹是一棵樹。2.樹中不存在迴路。3.森林不是樹。能夠用並查集判斷是否存在迴路，而後用並查集掃描一遍，看是否全部節點的最頂端父節點相同，不一樣則爲森林。

#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;

typedef struct
{
    int rank;
    int parent;
}UFSTree;

UFSTree Set[10000];
int Stack[10000];//用於保存每一個節點對中的父節點，以檢測是否爲森林

void MakeSet()
{
    for (int i = 0; i < 10000; i++)
    {
        Set[i].rank = 0;
        Set[i].parent = i;
    }
}

int FindSet(int x)
{
    if (x == Set[x].parent)
    {
        return x;
    }
    else
    {
        return FindSet(Set[x].parent);
    }
}

void UnionSet(int x, int y)
{
    x = FindSet(x);
    y = FindSet(y);
    if (Set[x].rank > Set[y].rank)
    {
        Set[y].parent = x;
    }
    else
    {
        Set[x].parent = y;
        if (Set[x].rank == Set[y].rank)
        {
            Set[y].rank++;
        }
    }
}

int main()
{
    int x, y, nCase = 1, top = 0;
    bool flag = true;
    MakeSet();
    while(scanf("%d%d", &x, &y) != NULL && x != -1 && y != -1)
    {
        while(1)
        {
            if (x == 0 && y == 0)
            {
                break;
            }
            Stack[top] = x;
            top++;
            //若是兩個節點處於同一集合，則不能造成一棵樹
            if (FindSet(x) == FindSet(y))
            {
                flag = false;
            }
            else
            {
                UnionSet(x, y);
            }
            scanf("%d%d", &x, &y);
        }
        if (flag)
        {
            //查詢每一個節點對中的父節點，看全部節點的最頂端父節點是否相同，不相同則爲森林
            for (int i = 0; i < top - 1; i++)
            {
                int n1 = FindSet(Stack[i]);
                int n2 = FindSet(Stack[i + 1]);
                if (n1 != n2)
                {
                    flag = false;
                    break;
                }
            }
        }
        if (flag)
        {
            flag = true;
            printf("Case %d is a tree.\n", nCase++);
        }
        else
        {
            flag = true;
            printf("Case %d is not a tree.\n", nCase++);
        }
        MakeSet();
        top = 0;
    }
    return 0;
}