Is it a Tree? 并查集 坑多

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.

这道题目一开始理解为就是是否成环，然而实际上...看了看讨论区这道题比较坑。

如果大家一直WA请测试以下用例吧.. 除了成环还有空树，森林，若简单地用点=边+1 第四个会WA等等..

0 0

1 1 0 0

1 2 1 2 0 0

1 2 2 3 3 1 4 4 1 0

1 2 2 3 3 1 0 0

只有0 0是tree 其他都不是。

AC代码：

#include <cstdio>using namespace std;int fa[100005];bool vis[100005];bool flag;void init(){    for(int i=1;i<=100000;i++)    {        fa[i]=i;        vis[i]=0;    }}int findd(int x){    if(fa[x]==x) return x;    else return fa[x]=findd(fa[x]);}void unite(int x,int y){    vis[x]=1;vis[y]=1;    int xx=findd(x), yy=findd(y);    if(xx==yy) flag=1;    else fa[xx]=yy;}int main(){    int i=1;    int n,m;    while(~scanf("%d%d",&n,&m))    {        if(n<0&&m<0) break;        if(!(n||m))        {            printf("Case %d is a tree.\n",i++);        }        else        {        flag=0;        init();        unite(n,m);        while(scanf("%d%d",&n,&m)&&n&&m) unite(n,m);        int num=0;        for(int j=1;j<=100000;j++)            if(vis[j]&&fa[j]==j) num++;        if(num>1) flag=1;        if(!flag) printf("Case %d is a tree.\n",i++);        else printf("Case %d is not a tree.\n",i++);        }    }    return 0;}

其实操作很简单，就是考察的比较细致..据说测试用例很水..