今天搞了一个下午，总算是把并查集了解了一点

杭电1325

Is It A Tree?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17660    Accepted Submission(s): 3969

Problem 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.附上代码：#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <iomanip>#include <cmath>using namespace std;#define maxn 10005int tree[maxn];//此为树的根int flag[maxn]；//这是标记数组int find(int x)//寻找父节点{    if(x==tree[x])    {        return x;    }    return tree[x]=find(tree[x]);}int main(){    int a,b,i,max;    bool p;    int cas=1;    while(scanf("%d%d",&a,&b)!=EOF)    {        max=0;        if(a<0 || b<0)        {            break;        }        p=1;//初始化，认为它可以是树        if(a==0 || b==0)        {            cout<<"Case "<<cas++<<" is a tree."<<endl;//空树也是树        }        for(i=1;i<=maxn;i++)        {            tree[i]=i;            flag[i]=0;        }        if(a==b)        {            p=0;        }else{            tree[b]=a;//令a为b的根        }        flag[a]=flag[b]=1;//对两个点进行标记        max=a>b?a:b;//寻找最大的那个，作用在后面几行你就明白了        while(scanf("%d%d",&a,&b)!=EOF)        {            if(a==0 || b==0)            {                break;            }            if(a==b)            {                p=0;            }            flag[a]=flag[b]=1;            max=max>a?max:a;            max=max>b?max:b;            if(tree[b]!=b || find(b)==find(a))//如果find（b）==find（a），那么说明已经组成环了            {                p=0;            }else{                tree[b]=a;            }        }        int num=0;        for(i=1;i<=max;i++)//我们从输入的最小数，到最大数进行遍历，这个好像可以减少时间        {            if(flag[i])//找到那些已经标记过的点            {                if(find(i)==i)//此为树的根的个数，因为树的根只能有一个                {                    num++;                    if(num>1)                    {                        p=0;                        break;                    }                }            }        }        if(num!=1)//如果根的个数大于一，那么说明不是树        {            p=0;        }        if(p)        {            cout<<"Case "<<cas++<<" is a tree."<<endl;        }else{            cout<<"Case "<<cas++<<" is not a tree."<<endl;        }    }    return 0;}