POJ1308 Is It A Tree?

Is It A Tree?

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 32054 Accepted: 10876

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 45 6  0 08 1  7 3  6 2  8 9  7 57 4  7 8  7 6  0 03 8  6 8  6 45 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

题目让判断所给的节点是否能够组成一个树，树的定义如下：首先空树也是树，其次，只能有一个根结点，也就是说不能是森林，不能存在入度大于1的节点或者从一个节点到另一个节点的路径不唯一，

#include <stdio.h>#include <string.h>using namespace std;int n,m,par[10005],vis[10005];void Init(){    for(int i = 1; i <= 10000; i ++){        par[i] = i;        vis[i] = 0;    }}int find_set(int x){    if(x == par[x])        return x;    return par[x] = find_set(par[x]);}void UnoinSet(int u, int v){    int x = find_set(u);    int y = find_set(v);    if(x != y)        par[y] = x;}int main(){    int ans = 0,flag;    while(~scanf("%d%d",&n,&m)){        if(n == -1 && m == -1)            break;        if(!n && !m){ //空树也是树            printf("Case %d is a tree.\n",++ans);            continue;        }        Init();        flag = 0;        int root = n;        if(n == m){//指向本身是不合法的            flag = 1;        }        else{            vis[n] = vis[m] = 1;//vis用来标记树中的节点            UnoinSet(n,m);        }        while(~scanf("%d%d",&n,&m)){            if(!n && !m)                break;            if(find_set(n) == find_set(m)){//会构成多条路径到一个节点或者有的节点入度大于1                flag = 1;            }            vis[n] = vis[m] = 1;            UnoinSet(n,m);        }        if(!flag){            for(int i = 1; i <= 10000; i ++){                if(vis[i] && find_set(i) != find_set(root)){//判断是否存在森林                    flag = 1;                    break;                }            }        }        if(flag)            printf("Case %d is not a tree.\n",++ans);        else            printf("Case %d is a tree.\n",++ans);    }    return 0;}