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.
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
2 5 3 1 2 2 3 4 5 5 1 2 5
Sample Output
2 4
给你一个数T，代表有几组数据，然后给你两个数m and n，表示有m个同学，
有n种关系，x 和 y 是同学，求最少需要多少桌子，
样例解释：
5 3
1 2
2 3
4 5
1和2是同学， 2和3是同学， 4和5是同学
所以只需要两张桌子，以为有两块（1，2，3）和（4，5）；
所以就有两块
解题思路：
纯的并查集，套模板就行，
直接上代码，
ps：我第一次是CE了，因为把下面的代码是AC的，CE是因为rank数组没用着还是咋回事，
我也不太清楚，有知道的可以告诉我哦，我只是把代码的一部分注释了就AC了，
/** 2015 - 09 - 18 晚上 Author: ITAK Motto: 今日的我要超越昨日的我，明日的我要胜过今日的我， 以创作出更好的代码为目标，不断地超越自己。 **/ #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <vector> #include <queue> #include <algorithm> #include <set> using namespace std; typedef long long LL; const int maxn = 30005; const double eps = 1e-7; struct node { int x, y; } arr[maxn]; int rank[maxn], fa[maxn], sum[maxn]; ///基本上是模板 void Init(int x) { for(int i=1; i<=x; i++) { fa[i] = i; ///rank[i] = 0; sum[i] = 1; } } int Find(int x) { if(fa[x] != x) fa[x] = Find(fa[x]); return fa[x]; } void Union(int x, int y) { int fx = Find(x), fy = Find(y); if(fx == fy) return; /** if(rank[fx] > rank[fy]) fa[fy] = fx, sum[fx] += sum[fy]; else { fa[fx] = fy; if(rank[fx] == rank[fy]) rank[fy]++; sum[fy] += sum[fx]; } **/ if(fx != fy) fa[fy] = fa[fx]; } int main() { int T, m, n, x, y; cin>>T; while(T--) { int cnt = 0; cin>>m>>n; Init(m); while(n--) { cin>>x>>y; Union(x, y); } for(int i=1; i<=m; i++) if(fa[i] == i) cnt++; cout<<cnt<<endl; } return 0; }
