询问给定图中树的棵数

Trees

Time Limit: 2500ms, Special Time Limit:6000ms, Memory Limit:32768KB

Problem 11348 : No special judgement

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.

Problem Source

  2012 Rocky Mountain Regional Contest

题意： 给定图中，树的颗数。

         图中可能有环，并查集，记录构成环的任意一点，最后利用此点处理环，可能有很高效的处理方法，但是这个数据规模这么做够了。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <map>
#include <vector>
#include <set>
#include <algorithm>
#include <vector>
#include <stack>
#define Max(a,b) ((a)>(b)?(a):(b))
using namespace std;
typedef long long LL ;
const int size=508 ;
int k=1 ;
struct Me{
    int N ;
    int M ;
    int father[size] ;
    vector<int>cycle ;
    Me(){} ;
    Me(int n,int m):N(n),M(m){
       for(int i=1;i<=N;i++)
          father[i]=i ;
       cycle.clear() ;
    }
    int find_father(int x){
       if(father[x]==x)
           return x ;
       else
           return father[x]=find_father(father[x]) ;
    }
    void Union(){
        int x ,y ,f=1;
        for(int i=1;i<=M;i++){
            scanf("%d%d",&x,&y) ;
            int f_x=find_father(x) ;
            int f_y=find_father(y) ;
            if(f_x==f_y)
               cycle.push_back(x) ;
            else
               father[f_x]=f_y ;
        }
        printf("Case %d: ",k++) ;
    }
    void gao(){
        this->Union() ;
        set<int>st ;
        set<int>cycle_father ;
        st.clear() ;
        cycle_father.clear() ;
        for(int i=0;i<cycle.size();i++)
            cycle_father.insert(find_father(cycle[i])) ;
        for(int i=1;i<=N;i++){
            int f=find_father(i) ;
            if(cycle_father.find(f)!=cycle_father.end())
                continue ;
            else
                st.insert(f) ;
        }
        if(st.size()==0)
           puts("No trees.") ;
        else  if(st.size()==1)
            printf("There is one tree.
") ;
        else
            printf("A forest of %d trees.
",st.size()) ;
    }
};
int main(){
   int n , m ;
   while(scanf("%d%d",&n,&m)){
       if(n==0&&m==0)
           break  ;
       Me me(n,m) ;
       me.gao() ;
   }
   return 0 ;
}