poj 1308 Is It A Tree?

http://poj.org/problem?id=1308

 Is It A Tree?

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

Submit Status

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 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.

和小希的迷宫很类似的一道题

判断树有四点：

1：判断是否只有一个根；

2：判断节点的总数是否比边的总数大一；

3：判断是否有回路；

4：判断每个节点的入度是否<=1;

之前自己按着这四点要求写了代码，发现好麻烦，最后看了别人的博客，发现根本不用那么麻烦，只要判断节点的总数是否比边的总数大一也就判断出了这个集合是否只有一个根，每个节点的入度是不是<=1；

解题思路：

输入两个数，判断是否在一个集合中，如果不在一个集合中，就把它们放到一个集合中；如果在一个集合中就说明这个集合有回路，返回false；还有计算总节点数和总边数

初次接触树的问题，花的时间可不少呢，不过还好，总算弄明白了

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <cmath>#include <cstdlib>#include <limits>#include <queue>#include <stack>#include <vector>#include <map>using namespace std;#define N 100010#define INF 0xfffffff#define PI acos (-1.0)#define EPS 1e-8int f[N], r[N], flag, flag1, flag2, flag3, flag4, vis[N];int Find (int x);//查找根节点void Init ();//初始化void join (int a, int b);//判断两个节点之间是否有回路int main (){    int a, b, k = 1;    while (scanf ("%d%d", &a, &b), a != -1 && b != -1)    {        flag = 1, flag1 = 0, flag2 = 0;        if (!a && !b)        {            cout << "Case " << k++ << " is a tree." << endl;            continue;        }        Init ();        while (a || b)        {            if (!vis[a]) flag1++, vis[a] = 1;            if (!vis[b]) flag1++, vis[b] = 1;            flag2++;            join (a, b);            scanf ("%d%d", &a, &b);        }        if (flag && flag1 == flag2 + 1)//没有回路并且节点数=边数+1            cout << "Case " << k++ << " is a tree." << endl;        else            cout << "Case " << k++ << " is not a tree." << endl;    }}int Find (int x){    if (x != f[x]) f[x] = Find (f[x]);    return f[x];}void join (int a, int b){    int x = Find (a), y = Find (b);    if (x != y) f[x] = y;//根节点合并    else flag = 0;}void Init (){    memset (vis, 0, sizeof (vis));    for (int i=1; i<=N; i++)        f[i] = i;}