HDU 1325POJ 1308 Is it A tree ? [并查集+树判定]

题目链接 poj hdu

Is It A Tree?

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 421 Accepted Submission(s): 157

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 lineCase 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.

程序分析（转载）：

　　这道题跟小希的迷宫有很大的相似吧，只是一个是无向图一个是有向图。也是给你那些结点之间的信息，然后让你判断是不是一颗树罢了，用树的定义来判断吧，无环，n个结点最多有n-1条边，不然就会有环。只有一个入度为0的结点，不存在入度大于1的结点。这些也足以判断是否为一棵树了吧。不过要注意一些特殊数据的情况，空树也是树。比如输入0 0。

解决方法：

　　其实也可以不用并查集，这样就可以直接按照上面的条件来统计，就可以判断是不是一颗树了。

自我感言：额，这道题目关键就是判断树的那几个标准，1，无环；2，除了根，所有的入度为1，根入度为0；3，这个结构只有一个根，不然是森林了。

这是三个标准拿捏好，再注意这里空树也是树。基本上就ok了。。

坑爹的是下边的代码可以在hdu上通过，而不能在poj上通过。。。。。看了discuss才知道 1 1 0 0 不能是树。

下面几种情况要注意了。。。。
1: 0 0 空树是一棵树
2: 1 1 0 0 不是树不能自己指向自己
3: 1 2 1 2 0 0 不是树….自己开始一直在这么WA 好郁闷重复都不行呀~~5555
4: 1 2 2 3 4 5 不是树森林不算是树(主要是注意自己)
5: 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 1 注意一个节点在指向自己的父亲或祖先都是错误的即 9–>1 错
6: 1 2 2 1 0 0 也是错误的这样基本上就可以在poj上通过了，poj(1308) 题这是hdu上通过的代码：

#include <stdio.h>const int max_num = 100000+10;
typedef struct
{int num,root,conn;//数据、根、入度
}Node;Node node[max_num];void init()
{for(int i = 0; i < max_num; i++){node[i].conn = 0;//入度初始化为0node[i].root= i;//根记录为自身node[i].num=0;//标记数字是否被使用过，0:没有被使用过，1:使用过了}
}int find_root(int a)
{if(node[a].root!=a)return node[a].root = find_root(node[a].root);return node[a].root;
}void union_set(int a,int b)
{a = find_root(a);b = find_root(b);if(a==b)//同一个根，说明是在同一个树下return;node[b].root=a;//把b的根赋为a的根，此时a已经是根，num==root
}int main()
{int n,m;int i = 1;bool flag=true;//true：是个树，false:不是树init();while(scanf("%d%d",&n,&m)!=EOF&&n>=0&&m>=0){if(!flag&&n!=0&&n!=0)continue;//已经确定不是树了，就继续循环if(n==0&&m==0){int root_num=0;for(int j = 1; j < max_num;j++){//判断是否为森林，如果，root_num用来记录根的数目if(node[j].num && find_root(j)==j)root_num++;if(node[j].conn>1)//如果出现某个节点的入度超过1，不是树{flag = false;break;}}if(root_num>1)//连通分支大于1，是森林不是树flag=false;if(flag)printf("Case %d is a tree.\n",i++);else printf("Case %d is not a tree.\n",i++);flag = true;init();continue;}if(m!=n&&find_root(n)==find_root(m))flag = false;else{//将m,n，记录为节点node[m].num = 1;node[n].num = 1;node[m].conn++;//入度增加一union_set(n,m);}}return 0;
}

中间只是稍作修改，把if(m!=n&&find_root(n)==find_root(m))改成了if((m!=n&&find_root(n)==find_root(m))||m==n）就在poj上通过了

#include <stdio.h>const int max_num = 100000+10;
typedef struct
{int num,root,conn;//数据、根、入度
}Node;Node node[max_num];void init()
{for(int i = 0; i < max_num; i++){node[i].conn = 0;//入度初始化为0node[i].root= i;//根记录为自身node[i].num=0;//标记数字是否被使用过，0:没有被使用过，1:使用过了}
}int find_root(int a)
{if(node[a].root!=a)return node[a].root = find_root(node[a].root);return node[a].root;
}void union_set(int a,int b)
{a = find_root(a);b = find_root(b);if(a==b)//同一个根，说明是在同一个树下return;node[b].root=a;//把b的根赋为a的根，此时a已经是根，num==root
}int main()
{int n,m;int i = 1;bool flag=true;//true：是个树，false:不是树init();while(scanf("%d%d",&n,&m)!=EOF&&n>=0&&m>=0){if(!flag&&n!=0&&n!=0)continue;//已经确定不是树了，就继续循环if(n==0&&m==0){int root_num=0;for(int j = 1; j < max_num;j++){//判断是否为森林，如果，root_num用来记录根的数目if(node[j].num && find_root(j)==j)root_num++;if(node[j].conn>1)//如果出现某个节点的入度超过1，不是树{flag = false;break;}}if(root_num>1)//连通分支大于1，是森林不是树flag=false;if(flag)printf("Case %d is a tree.\n",i++);else printf("Case %d is not a tree.\n",i++);flag = true;init();continue;}if((m!=n&&find_root(n)==find_root(m))||m==n)flag = false;else{//将m,n，记录为节点node[m].num = 1;node[n].num = 1;node[m].conn++;//入度增加一union_set(n,m);}}return 0;
}

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