hnsd11348tree(并查集)

Problem description

A graph consists of a set of vertices and edges between pairs of vertices. Two vertices are connected if there is a path (subset of edges) leading from one vertex to another, and a connected component is a maximal subset of vertices that are all connected to each other. A graph consists of one or more connected components.

A tree is a connected component without cycles, but it can also be characterized in other ways. For example, a tree consisting of n vertices has exactly n-1 edges. Also, there is a unique path connecting any pair of vertices in a tree.

Given a graph, report the number of connected components that are also trees.

Input

The input consists of a number of cases. Each case starts with two non-negative integers n and m, satisfying n ≤ 500 and m ≤ n(n-1)/2. This is followed by m lines, each containing two integers specifying the two distinct vertices connected by an edge. No edge will be specified twice (or given again in a different order). The vertices are labelled 1 to n. The end of input is indicated by a line containing n = m = 0.

Output

For each case, print one of the following lines depending on how many different connected components are trees (T > 1 below):

	Case x: A forest of T trees.
	Case x: There is one tree.
	Case x: No trees.
      

x is the case number (starting from 1).

Sample Input

6 3
1 2
2 3
3 4
6 5
1 2
2 3
3 4
4 5
5 6
6 6
1 2
2 3
1 3
4 5
5 6
6 4
0 0

Sample Output

Case 1: A forest of 3 trees.
Case 2: There is one tree.
Case 3: No trees.

#include<stdio.h>
int fath[505],cycl[505],k,n;
void setfirst()
{
    k=n;
    for(int i=1;i<=n;i++)
    {
        fath[i]=i; cycl[i]=0;
    }
}
int find_fath(int x)
{
    if(x!=fath[x])
    fath[x]=find_fath(fath[x]);
    return fath[x];
}
void setTree(int a,int b)
{
    a=find_fath(a);
    b=find_fath(b);
    if(cycl[b]&&cycl[a])
    return ;
    k--;
    if(a!=b)
    {
        if(cycl[a])
        fath[b]=a;
        else
        fath[a]=b;
    }
    else
    cycl[a]=1;
}
int main()
{
    int a,b,m,t=1;
    while(scanf("%d%d",&n,&m)>0&&m+n!=0)
    {
        setfirst();
        while(m--)
        {
            scanf("%d%d",&a,&b);
            setTree(a,b);
        }
        printf("Case %d: ",t++);
        if(k>1)printf("A forest of %d trees.
",k);
        if(k==1)printf("There is one tree.
");
        if(k==0)printf("No trees.
");
    }
}