HDU 1325 有根树的判断

Is It A Tree?

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 33083 Accepted Submission(s): 7574

题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=1325

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.

Source

North Central North America 1997

题意

给你一些有向边，问你这些点是否构成一颗有根树。

题解

有根树特点：

１．除根点入度为０外，其余点入度均为１

２．没有环

无根树特点：

１．没有环

２．联通

奇环内向数特点：

１．入度全为１

２．联通

我们要做的是有根树，那么用并查集处理环，用set或者map处理每个点的入度即可。

代码

#include<bits/stdc++.h>
using namespace std;
#define N 100050
int find(int fa[],int x)
{
    if (x==fa[x])return x;
    return fa[x]=find(fa,fa[x]);
}
bool pd()
{
    static int f[N],bl[N];
    set<int>Q;
    memset(f,0,sizeof(f));
    for(int i=1;i<N;i++)bl[i]=i;
    int num=0,x,y,flag=1;
    while(1)
    {
        scanf("%d %d",&x,&y);
        if (x+y==0)break;
        if (x<0)exit(0);
        Q.insert(x);
        Q.insert(y);
        f[y]++;
        int fx=find(bl,x),fy=find(bl,y);
        if  (fx==fy||f[y]>1)flag=0;
        bl[fx]=fy;
    }
    for(auto it=Q.begin();it!=Q.end();it++) if (f[*it]==0)num++;
    if (flag&&num<=1)return 1;
    return 0;
}
void work()
{
    static int cas=0;
    bool k=pd();
    printf("Case %d is ",++cas);
    if (!k)printf("not ");
    printf("a tree.\n");
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("aa.in","r",stdin);
#endif
    while(1)work();
}