HDU - 1325 Is It A Tree?

Is It A Tree?

Time Limit: 1000MS Memory Limit: 32768KB 64bit IO Format: %I64d & %I64u

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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.        

 

Source

North Central North America 1997

本题与hdu1272小希的迷宫几乎一样，区别是这个题是有向图，而小希的迷宫是无向图。另外，有点坑的是它的输入结束标志是一对负数，而不是-1（为此多次Time Limit Exceeded）。代码与小希的迷宫几乎一样。代码：

#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=100005;
int fa[maxn],sign[maxn];//多了sign[],用于统计每个节点的入度（作为线段终点的次数，即作为儿子的次数。显然每个点只能有一个父亲，所以入度不能大于1）
bool vis[maxn];
int find(int x)
{
    while(x!=fa[x])
    x=fa[x];
    return x;
}
int main()
{
    int n,m,i,a,b,kase=0;
    while(scanf("%d%d",&n,&m)&&n>=0&&m>0)//这里输入结束标志是一对负数
    {
        kase++;
        memset(vis,false,sizeof(vis));
        memset(sign,0,sizeof(sign));//初始化入度都是0
        bool flag=true;
        for(i=1;i<=maxn;i++)
        fa[i]=i;
        fa[m]=n;
        vis[n]=vis[m]=true;
        sign[m]++;
        if(n==0&&m==0) {printf("Case %d is a tree.\n",kase);continue;}
        while(scanf("%d%d",&n,&m)&&n!=0&&m!=0)
        {
            vis[n]=vis[m]=true;
            sign[m]++;
            a=find(n);
            b=find(m);
            if(a!=b)
            fa[b]=a;
            else flag=false;
        }
            int k=0;
            for(i=1;i<maxn;i++)
            {
                if(fa[i]==i&&vis[i])
                k++;
                if(sign[i]>1) flag=false;//入度不能大于1
            }
           if(k>1) flag=false;
           if(flag) printf("Case %d is a tree.\n",kase);
           else printf("Case %d is not a tree.\n",kase);
    }
    return 0;
}

总结：

有向图与无向图的区别。对于这个题其实就一个区别：入度。