poj 1308(并查集)

题目：

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.

题解：这道题就是判断给出一系列点对，看是否能组成一个树，注意点就是判断的情况要全，这里给出几组特殊数据。

0 0 空树也是一棵树

1 1 0 0  不是树 不能自己指向自己

1 2 1 2 0 0 不是树 不可以重复

1 2 2 3 4 5 不是树 有多个根节点也就是有多颗数不可以

1 1 2 2 3 3 4 4 5 1 不是树 子节点不可以是根节点

#include <stdio.h>
const int N = 1005;
int m,n,pa[N],flag1,cases = 1,k,flag2;
int mapp[N];//记录出现的编号
void init(){
	for(int i = 1; i <= N-1 ; i++){
		pa[i] = i;
		mapp[i] = 0;
	}
	flag1 = 1;
	flag2 = 1;
}
int get_parent(int x){
    return pa[x] == x ? x : get_parent(pa[x]);
}
void merge(int x,int y){
    mapp[x] = 1;//编号出现记为1
    mapp[y] = 1;
	int px = get_parent(x);
	int py = get_parent(y);
	if(px != py)
        pa[py] = pa[px];
}
void judge(int x,int y){
	if(pa[y] != y)//判断子节点是否已经有父节点
		flag1 = 0;
    else
        if(get_parent(x) == y)//判断子节点是否为根节点
            flag1 = 0;
}
int main(){
    while(1){
        init();
        int i,j;
        while(scanf("%d%d",&m,&n) != EOF){
            if(m < 0 && n < 0)
                return 0;
            if(m == 0 && n == 0){
                if(k == 0)//记录是不是空树
                    printf("Case %d is a tree.\n", cases++);//空树也可以
                break;
            }
            k++;
            if(m == n)//两个结点不能相等
                flag2 = 0;
            judge(m,n);
            if(flag1){
                 merge(m,n);
                 flag1 = 1;
            }
            else
                flag2 = 0;
        }
        if(k == 0)
            continue;
        for(i = 1;i <= N-1;i++)//得到根节点
            if(mapp[i] == 1){
                j = get_parent(i);
                break;
            }
        for(int q = i+1 ;q <= N-1; q++)//判断是否有多个根节点，即是否只有一个树
                if(mapp[q] == 1 && j != get_parent(q)){
                    flag2 = 0;
                    break;
                }
        if(flag2)
            printf("Case %d is a tree.\n",cases++);
        else
            printf("Case %d is not a tree.\n",cases++);
        k = 0;//标记还原
    }
	return 0;
}