OpenJudge noi 310 Is It A tree（POJ1308）

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.

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 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
样例输出
Case 1 is a tree.
Case 2 is a tree.
Case 3 is not a tree.

这个题我们可以直接判断每个点入度，首先需要保证的是入度必须<=1,这个很显然。然后我们要特判空树（没有元素也是一个树）和森林的情况（入度等于0的点>1）.

代码如下：

#include<iostream>#include<cstdio>#include<cstdlib>using namespace std;const int maxn=100005;int du[maxn],a[maxn];bool sss[maxn];int main(){    bool flag;    int n,m,ss=0,tot=0;    while(true)    {        scanf("%d%d",&n,&m);        if(n==-1&&m==-1)            break;              if(n==0&&m==0)        {            flag=0;            ss++;            int cnt=0;                      if(tot==0)                printf("Case %d is a tree.\n",ss);            else            {                for(int i=1;i<=tot;i++)                {            //      cout<<a[i]<<" a[i] "<<du[a[i]]<<" du "<<endl;                    if(du[a[i]]==0)                    {                        cnt++;                        if(cnt>1)                            flag=1;                             }                    if(du[a[i]]>1)                        flag=1;                    du[a[i]]=0;                }                 if(flag==1)                    printf("Case %d is not a tree.\n",ss);                else if(flag==0)                    printf("Case %d is a tree.\n",ss);                for(int i=1;i<tot;i++)                      sss[a[i]]=0;                tot=0;                                      }                   }        if(sss[n]==0)        {            sss[n]=1;            a[++tot]=n;                 }        if(sss[m]==0)        {            sss[m]=1;            a[++tot]=m;                 }        du[m]++;    }}

这个题目也可以用并查集来解决。就是说，由题目可知，我们可以处理出来每个点的父亲，输入结束时，我们去判断一下他们是不是同属于一个并查集。当然这个也需要特判空树的情况。还有就是如果重复建边，也是不正确的。（对于一棵树，n点n-1条边，重复建边显然是不会满足树的性质的），大体思路就是这样，代码就不放了。