hdu 1325 Is It A Tree? （无向图）

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. 

InputThe 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. 
OutputFor 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.

题目意思：给你一些结点之间的信息，然后让你判断是不是一颗树。
直接用树的定义来做，无环，n个结点有n-1条边就是树。

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
int vis[100005];
int main()
{
    int n,m,flag,i,j;
    int t=0;
    while(1)
    {
        i=j=flag=0;//j记录有几个点，i记录有几条边 
        memset(vis,0,sizeof(vis));
        while(scanf("%d%d",&n,&m)!=EOF&&n&&m)
        {
            if(n<0||m<0)
                return 0;
            if(vis[m]==2)//m有两个以上入度（有环)则不是树 
                flag=1;
            if(vis[n]==0)
                j++;
            if(vis[m]==0)
                j++;
            vis[n]=1;vis[m]=2;i++;//用于下次n,m所以n,m入度变为1,2 
        }
        if(flag==0&&j==i+1)//无环且点==边+1则为树 
            printf("Case %d is a tree.
",++t);
        else
            printf("Case %d is not a tree.
",++t);
    }
    return 0;
}