撰写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.
InputThe 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.
OutputFor 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
直接贴代码：
#include<algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <vector>
#include <string>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define P(a,b,c) make_pair(a,make_pair(b,c))
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n;i>=a;i--)
#define CLR(vis) memset(vis,0,sizeof(vis))
#define MST(vis,pos) memset(vis,pos,sizeof(vis))
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef pair<int,pair<int,int> >pii;
typedef long long ll;
const ll mod = 1000000007;
ll gcd(ll a, ll b) { return b ? gcd(b, a%b) : a; }
const int maxn = 3e4+5;
int pre[maxn];
int rankk[maxn];
int find(int x){
return pre[x]==x?x:pre[x]=find(pre[x]);
}
void unionn(int a,int b){
a = find(a);
b = find(b);
if(a==b)return;
else{
if(rankk[a]<rankk[b]) pre[a]=b;
else {
pre[b] = a;
if(rankk[a] == rankk[b]) rankk[a]++;
}
}
}
void init(int n){
rep(i,1,n){
rankk[i]=1;
pre[i]=i;
}
}
set<int>s;
int main(){
int n,m,t;
scanf("%d", &t);
while(t--){
s.clear();
scanf("%d%d", &n,&m);
init(n);
while(m--){
int u,v;
scanf("%d%d",&u,&v);
unionn(u,v);
}
int cnt = 0;
rep(i,1,n){
if(!s.count(find(i))){
s.insert(find(i));
cnt++;
}
}
cout<<cnt<<endl;
}
return 0;
}