POJ 1308-Is It A Tree?（並查集）

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.

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.

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).

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

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

North Central North America 1997

#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <iomanip>
#include <map>
#include <ctime>
#define INF 0x3f3f3f3f
#define MAXN 110
using namespace std;
int fa[MAXN];
bool vis[MAXN];
int fa_find(int p)
{
    if(fa[p]==p) return p;
    return fa[p]=fa_find(fa[p]);
}
bool join(int p,int q)
{
    int p1=fa_find(p);
    int q1=fa_find(q);
    if(q1!=q)  return true;//q不是根節點
    if(p1==q1) return true;//p和q已經在同一棵樹
    else fa[q1]=p1;//合併兩棵樹
    return false;
}
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("G:/cbx/read.txt","r",stdin);
    //freopen("G:/cbx/out.txt","w",stdout);
#endif
    int n=0,a,b,ca=0,ans=0;
    for(int i=0; i<MAXN; ++i) fa[i]=i;
    memset(vis,false,sizeof(vis));
    while(~scanf("%d%d",&a,&b))
    {
        if(a==-1&&b==-1) break;
        if(a==0&&b==0)
        {
            int cnt=0;
            for(int i=0; i<n; i++)
            {
                if(fa[i]==i&&vis[i])//在出現過的節點中統計樹根的數量
                    ++cnt;
                if(cnt>1) break;
            }
            if(cnt>1||ans>0)//多棵樹或者出現過不合法的情況
                cout<<"Case "<<++ca<<" is not a tree."<<endl;
            else cout<<"Case "<<++ca<<" is a tree."<<endl;
            ans=0;//初始化
            for(int i=0; i<MAXN; ++i) fa[i]=i;
            memset(vis,false,sizeof(vis));
            continue;
        }
        n=max(n,a);
        n=max(n,b);//記錄最大的序號
        vis[a]=vis[b]=true;//標記節點出現過
        if(join(a,b)) ++ans;//出現不合法情況
    }
    return 0;
}