HDU1325 Is It A Tree? （并查集判断有向图是否是树）

Is It A Tree?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 36235    Accepted Submission(s): 8310

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 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

 

Sample Output

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

 

题意与分析

本题和HDU1272几乎一模一样，唯一的区别是本题是有向图

所以在使用并查集的时候要注意判断一下点的入度最多为1

也可以用拓扑排序

或者直接按树的定义做 但是这两种方法并没有AC。。。。

#include<bits/stdc++.h>
using namespace std;

int s[100005],a,b,flag,root,vis[100005],cases=0;

int findf(int x)
{
    return x==s[x]?x:s[x]=findf(s[x]);
}

void hebing(int a,int b)
{
    int fa=findf(a);
    int fb=findf(b);
    if(fa!=fb)
    {
        s[fb]=fa;
    }
}

int main()
{
    for(int i=1;i<=100000;i++)
    {
        s[i]=i;
        vis[i]=0;
    }
    while(~scanf("%d%d",&a,&b))
    {
        if(a==-1&&b==-1)
        {
            return 0;
        }
        else if(a+b==0)
        {
            for(int i=1;i<=100000;i++)
            {
                if(vis[i]==1&&s[i]==i)
                root++;          //统计根节点数量 
                if(root>1)
                {
                    flag=1;
                    break;
                }
            }
            if(flag)
            {
                cases++; 
                printf("Case %d is not a tree.
",cases);
            }
            else
            {
                cases++; 
                printf("Case %d is a tree.
",cases);
            }
            flag=0;
            root=0;
            for(int i=1;i<=100000;i++)
            {
                s[i]=i;
                vis[i]=0;
            }
        }
        else
        {
            if(vis[b]==1)
            {
                flag=1;
            }
            else
            {
                vis[a]=1;
                vis[b]=1;
                int fa=findf(a);
                int fb=findf(b);
                if(fa==fb)
                {
                    flag=1;
                }
                else
                hebing(a,b);
            }
        }
    }
}