HDU 1325 Is It A Tree?

HDU 1325 Is It A Tree?

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.

Source North Central North America 1997

樹的判斷，主要有三點

1：不能有環，

2：不能有兩個獨立的區間，為此我開了一個vis[N],來標記出現的數字

3：一個點的入度不能大於1，意思就是 a->b c->b,那麼b的入度就大於1了

注意的地方應該都說完了，上代碼喽(⊙o⊙)…

#include
#include
#include
#include
using namespace std;

#define N 1005

int father[N],vis[N],d[N];
int n,m;


int cha(int x)
{
    if(x!=father[x])
        father[x]=cha(father[x]);
    return father[x];
}

int main()
{
    int i,a,b,ca=0;

    while(scanf("%d%d",&a,&b))
    {
        if(a<0&&b<0) break;

        if(a==0&&b==0)  //空樹也是樹
        {
            printf("Case %d is a tree.\n",++ca);
            continue;
        }

        for(i=0;i1)  flag=1; //單獨點的入度不能大於2

               if(num>1)   flag=1;

               if(flag)    break;
           }

               if(flag)
                 printf("Case %d is not a tree.\n",++ca);
               else
                 printf("Case %d is a tree.\n",++ca);
       }
    }
    return 0;
}