HDOJ 1325 Is It A Tree?（并查集+入度判断）

Is It A Tree?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23572    Accepted Submission(s): 5386

Problem 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.
  
  

 

被一个等号搞得WA了半天，题目测试数据偏弱，不足请指正

#include <iostream>
#include <string.h>
using namespace std;
const int MAX_N=105;//数组不用开太大，105就够了
int par[MAX_N],num[MAX_N];//num数组为记录每个结点的入度
int x,y;
bool vis[MAX_N];
bool flag;
void init()//并查集初始化
{
	memset(vis,0,sizeof(vis));
	memset(num,0,sizeof(num));
	for(int i=0;i<MAX_N;i++)
	  	par[i]=i;
}
int find(int x)
{
	return par[x]==x?x:(par[x]=find(par[x]));
}
void unit(int x,int y)
{
	x=find(x);
	y=find(y);
	if(x==y)
	return;
	else
	par[x]=y;
}
bool same(int x,int y)
{
	return find(x)==find(y);
}
int main()
{
	int k=1;
	while(cin>>x>>y&&(x>-1&&y>-1))//这里要注意，当x和y小于0的时候结束，而不是x和y等于-1
	{
		init();
		flag=true;
		if(x==0&&y==0)//x==0&&y==0时是一棵树，即空集
		{
			printf("Case %d is a tree.\n",k++);
			continue;
		}
		vis[x]=vis[y]=true;//记录点是否使用过
		if(same(x,y))//比如x=1，y=1，则是一个环，不是树
		flag=false;
		else
		{
		unit(x,y);
		num[y]++;//y的入度+1
		}
		while(cin>>x>>y&&(x!=0&&y!=0))
		{
			vis[x]=vis[y]=true;
			if(same(x,y))
			flag=false;
			else
			{
			unit(x,y);
			num[y]++;
			}
		}
		int sum=0;
		for(int i=1;i<MAX_N;i++)
		{
			if(vis[i]&&par[i]==i)//判断是否是森林
				sum++;
			if(num[i]>1)//若有结点的入度大于1的时候，则不是树
			{
				flag=false;
				break;
			}
		}
		if(sum>1)
		flag=false;
		if(flag)
		printf("Case %d is a tree.\n",k++);
		else
		printf("Case %d is not a tree.\n",k++);
	}
	return 0;
}