hdu 1325 Is It A Tree?

 

Is It A Tree?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14276    Accepted Submission(s): 3188

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.

 

 

这道题跟小希的迷宫有很大的相似吧，只是一个是无向图一个是有向图。也是给你那些结点之间的信息，然后让你判断是不是一颗树罢了，用树的定义来 判断吧，无环，n个结点最多有n-1条边，不然就会有环。只有一个入度为0的结点，不存在入度大于1的结点。这些也足以判断是否为一棵树了吧。不过要注意 一些特殊数据的情况，空树也是树。比如输入0 0。

解决方法：

　　其实也可以不用并查集，这样就可以直接按照上面的条件来统计，就可以判断是不是一颗树了。

自我感言：额，这道题目关键就是判断树的那几个标准，1，无环；2，除了根，所有的入度为1，根入度为0；3，这个结构只有一个根，不然是森林了。

下面几种情况要注意了。。。。

1: 0 0 空树是一棵树

2: 1 1 0 0 不是树 不能自己指向自己

3: 1 2 1 2 0 0 不是树....

4: 1 2 2 3 4 5 不是树  森林不算是树(主要是注意自己)

5: 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 1  注意 一个节点在指向自己的父亲或祖先 都是错误的 即 9-->1 错

6: 1 2 2 1 0 0 也是错误的

 

#include<cstdio>

#include<cstring>

#include<iostream>

#include<string>

#include<algorithm>

using namespace std;

//const int maxn=100000+5;

const int maxn=1000+5;

int p[maxn], rank[maxn], indegree[maxn];

int used[maxn];



void make_set() {

    memset(p, -1, sizeof(p));

    memset(indegree, 0, sizeof(indegree));

    memset(rank, 0, sizeof(rank));

    memset(used, 0, sizeof(used));

}

int find_set(int x) {

    return p[x]==-1 ? x : p[x]=find_set(p[x]);

}



void union_set(int x, int y)

{

    int fx=find_set(x), fy=find_set(y);

    if(fx==fy) return;

    if(rank[fx] < rank[fy])

        p[fx]=fy;

    else {

        p[fy]=fx;

        if(rank[fx]==rank[fy]) rank[fx]++;

    }

}





//判断是否有回路

//判断入度是否有大于1的

//判断图是否连通

int main()

{

    int kase=0;

    int x,y;

    bool have_data;

    make_set(); 

    have_data=false;

    bool ok=true;

    while(scanf("%d%d", &x, &y)!=EOF && !(x==-1 && y==-1)) {

        if(x==0 && y==0)

        {

            //计算连通分量个数

            int cnt=0;

            for(int i=0;i<maxn;i++) if(used[i] && p[i]==-1) cnt++;

            if(cnt!=1) ok=false;



            //0 0情况，没有数据也是yes的

            if(!have_data || ok) printf("Case %d is a tree.\n", ++kase);

            else printf("Case %d is not a tree.\n", ++kase);



            //重新初始化

            make_set(); 

            have_data=false;

            ok=true;

            continue;

        }

        

        if(!ok) continue;

        //判断是否有回路

        if(find_set(x)==find_set(y))

        {

            ok=false; 

        }

        union_set(x, y);

        used[x]=used[y]=1;

        have_data=true;

        indegree[y]++;

        //判断入度是否有大于1的

        if(indegree[y]>1)

            ok=false;

    }

    return 0;

}