HDU 1213 How many tables

和HDU1232的畅通工程基本一模一样；

但是就是不需要ans-1；

题中问有多少个桌子，如果是连同的则为一个桌子，否则单独算；

并查集的找集合个数典型问题；

PS：学长的递归方法写查询函数还是简单的多；

How Many Tables

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21783    Accepted Submission(s): 10791

Problem Description

Today is Ignatius' birthday. He invites a lot of friends. Now it's dinner time. Ignatius wants to know how many tables he needs at least. You have to notice that not all the friends know each other, and all the friends do not want to stay with strangers.

One important rule for this problem is that if I tell you A knows B, and B knows C, that means A, B, C know each other, so they can stay in one table.

For example: If I tell you A knows B, B knows C, and D knows E, so A, B, C can stay in one table, and D, E have to stay in the other one. So Ignatius needs 2 tables at least.

 

 

Input

The input starts with an integer T(1<=T<=25) which indicate the number of test cases. Then T test cases follow. Each test case starts with two integers N and M(1<=N,M<=1000). N indicates the number of friends, the friends are marked from 1 to N. Then M lines follow. Each line consists of two integers A and B(A!=B), that means friend A and friend B know each other. There will be a blank line between two cases.

 

 

Output

For each test case, just output how many tables Ignatius needs at least. Do NOT print any blanks.

 

 

Sample Input

25 31 22 34 55 12 5

 

 

Sample Output

24

AC代码：

#include <bits/stdc++.h>using namespace std ;int pre[1111],vis[1111];int find(int x ){if(x!=pre[x])return pre[x]=find(pre[x]);return x ;}void mix (int x , int y ){int dx =find(x);int dy =find(y);if(dx!=dy){pre[dx]=dy;}}int main(){int t,a ,b ,ans , n , m  ;cin>>t;while(t--){ans = 0 ;memset(vis,0,sizeof(vis));cin>>n>>m;for(int i = 1 ; i<=n;i++){pre[i]=i;}for(int i =1 ;i<=m;i++){cin>>a>>b;mix(a,b);}for(int i =1 ; i<=n;i++){vis[find(i)]=1;}for(int i =1 ;i<=n;i++){if(vis[i]){ans++;}}printf("%d\n",ans);}return 0;}