Is It A Tree? POJ - 1308

问题：

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.

大意：

判断给出的这些点是否能形成一棵有向树，以0 0表示结尾，-1，-1表示结束程序。

思路：

1：对于有向树，根节点只有一个，根节点的入度必须为0，其他点的入度为1。

2：因为点的顺序不是连续的所以，所以要标记一下出现过的点。

3：树中不能存在环。

还有就是当只输入0 0时，视为可以形成树。

代码：

#define N 100010
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int f[N],k[N],r[N],c[N];
int getf(int a)
{
    if(f[a]==a)return a;
    return f[a]=getf(f[a]);
}
int main()
{
    int a,b,Case=1;
    while(scanf("%d%d",&a,&b)&&(a!=-1||b!=-1))
    {
        if(a==0&&b==0)
        {
            printf("Case %d is a tree.\n",Case++);
            continue;
        }
        for(int i=1; i<=N; i++)
            f[i]=i;
        memset(k,0,sizeof(k));
        memset(r,0,sizeof(r));
        int flag=1;
        do
        {
            k[a]=1;
            k[b]=1;
            r[b]++;
            int t1=getf(a);
            int t2=getf(b);
            if(t1==t2||a==b||r[b]>1)
                flag=0;
            else
                f[t2]=t1;
        }
        while(scanf("%d%d",&a,&b)&&(a||b));
        int sum=0,fsum=0;
        for(int i=1; i<=N; i++)
        {
            if(k[i]==1&&f[i]==i)
                sum++;
            if(k[i]==1&&r[i]==0)
                fsum++;
        }
        if(sum!=1||fsum!=1)flag=0;
        if(flag)printf("Case %d is a tree.\n",Case++);
        else  printf("Case %d is not a tree.\n",Case++);
    }
    return 0;
}