NYOJ129树的判定

树的判定

时间限制：1000 ms  |  内存限制：65535 KB

难度：4

描述

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.

输入

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.

The number of test cases will not more than 20,and the number of the node will not exceed 10000.
The inputs will be ended by a pair of -1.

输出

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).

样例输入

6 8  5 3  5 2  6 4 5 6  0 08 1  7 3  6 2  8 9  7 5 7 4  7 8  7 6  0 03 8  6 8  6 4 5 3  5 6  5 2  0 0-1 -1

样例输出

Case 1 is a tree.Case 2 is a tree.Case 3 is not a tree.

#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0        node[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;}

http://www.cnblogs.com/newpanderking/archive/2012/10/14/2723260.html

#include<stdio.h>  #include<string.h>  #define N 1000  int p[N]={0};  int find(int x)  {      //return x==p[x]?x:p[x]=find(p[x]);      while(x!=p[x])      x=p[x];      return x;  }  int fun(int x,int y)  {      int fy,fx;      fy=find(y);      fx=find(x);      if(fy!=fx)      {          p[fy]=fx;      }  }  int main()  {      int k,i,a,b,flag;      int sum;      k=1;      while(1)      {          flag=0;          while(scanf("%d %d",&a,&b)!=EOF&&a&&b)          {              if(a==-1||b==-1)              return 0;              if(p[a]==0) p[a]=a;              if(p[b]==0) p[b]=b;              if(find(a)==find(b)) flag=1;              if(flag!=1)              fun(a,b);          }          for(i=1,sum=0;i<=N;i++)          {              if(p[i]==i) sum++;              p[i]=0;          }          if(sum>1||flag==1)          printf("Case %d is not a tree.\n",k);          else printf("Case %d is a tree.\n",k);          k++;      }      return 0;  }  

http://blog.csdn.net/sungaochao/article/details/41380989