并查集判树的个数

链接：https://ac.nowcoder.com/acm/contest/12794/F
来源：牛客网

After two years of sharing a bedroom with you in a college dorm, Jeff finally has his own room. Excited about inviting girls over to his room, he ponders over what decorations the fairer sex will enjoy.¹ He decides upon setting up a “fake” planetarium with a black ceiling and glow in the dark stars that form constellations. Unfortunately, in his haste, he has made several “errors” in setting up his constellations. See, everyone knows that constellations don’t have cycles in them. Instead, whenever we visually connect the stars together with lines, a tree is always formed. (A cycle is formed when you can start at a star and, using connections, go to one or more other stars and then end up at the original star.)

Since you were Jeff’s roommate for two years, you figure you’ll help him fix his constellations. Your job will be twofold: to count the number of constellations Jeff has, and to report how many of them have cycles and need to be fixed. A constellation consists of multiple stars that are all connected to one another (directly or indirectly). A constellation that needs fixing is simply one that has a cycle.

 

The Problem:

 

Given several configurations of stars and connections between stars, determine how many constellations are defined in each configuration and how many need fixing.

输入描述:

The first input line contains a positive integer, n (n ≤ 100), indicating the number of night skies to  consider.   The  first  line  of  each  night  sky  contains  two positive  integers,  s  (s ≤  1000), representing  the  number  of  stars  for  this  night  sky,  and  c  (c  ≤  10000),  representing  the  total number  of  connections  between  pairs  of  stars  for  this  night  sky.  Each  of the following c input lines  contains  two  distinct positive integers representing a single connection between two stars. The stars in each test case will be numbered 1 through s, inclusive.  A connection is considered bidirectional, thus, if a is connected to b, b is connected to a.  Assume that all connections in a data set are distinct, i.e., no duplicates.  Also assume that there will never be a connection from a star to itself.

1 This is based on a true story.  The person who is the inspiration for this story did, in fact, major in Planetary Science and make his room’s ceiling a relatively accurate depiction of the night sky, as seen in Boston circa 1995.

输出描述:

For each test case, just output a line with the following format:
Night sky #k: X constellations, of which Y need to be fixed.
where  k is  the  number  of  the  night  sky,  starting  at  1,  X is  the  total  number  of  constellations described in that night sky, and Y is how many of those constellations contain a cycle.
Leave a blank line after the output for each test case. 

 

示例1

输入

复制 3 5 4 1 2 1 3 2 3 4 5 8 5 1 2 3 4 6 7 6 8 8 7 3 2 1 2 1 3

3
5 4
1 2
1 3
2 3
4 5
8 5
1 2
3 4
6 7
6 8
8 7
3 2
1 2
1 3

输出

复制 Night sky #1: 2 constellations, of which 1 need to be fixed. Night sky #2: 3 constellations, of which 1 need to be fixed. Night sky #3: 1 constellations, of which 0 need to be fixed.

Night sky #1: 2 constellations, of which 1 need to be fixed. 

Night sky #2: 3 constellations, of which 1 need to be fixed. 

Night sky #3: 1 constellations, of which 0 need to be fixed.

说明

Note:In the second example, star number 5 is not connected to any other stars. This staron its own is NOT counted as a constellation, leaving only {1,2}, {3,4}and {6,7,8} as constellations, of which only the last one needs tobe fixed.

 

 

 

就是判断这个无向图中，有多少个子树，和环的个数（就是这个环如果出现在一颗树上，算一个环），特别的时如果只有一个点

不算一个树

#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=1e6+100;
int r[maxn];    
int n,m;
int pre[maxn];
int vis[maxn];
int find(int x){
    if(x==pre[x]){
        return x;
    }
    else{
        return pre[x]=find(pre[x]);
    }
}
int main(){
    int t;
    int kase=0;
    cin>>t;
    while(t--){
        cin>>n>>m;
        for(int i=1;i<=n;i++){
            r[i]=1;
            vis[i]=0;
            pre[i]=i;
        }
        int x,y;
        for(int i=1;i<=m;i++){
            cin>>x>>y;
            int fx=find(x);
            int fy=find(y);
            if(fx!=fy){
                pre[fx]=fy;
                r[fy]+=r[fx];
                if(vis[fx]||vis[fy]){
                    vis[fx]=1;
                    vis[fy]=1;
                }
            }
            else{
                vis[fx]=1;
            }
        }
        int ans1=0;
        int ans2=0;
        for(int i=1;i<=n;i++){
            if(pre[i]==i&&r[i]>1){
                ans1++;
            }
            if(pre[i]==i&&vis[i]){
                ans2++;
            }
        }
        printf("Night sky #%d: %d constellations, of which %d need to be fixed.\n",++kase,ans1,ans2);
        puts("");
    }
}