C

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

解题思路：并查集的类型都相差不大，所以我仍然用的模板。只是稍作修改

#include <iostream>#include <cstdio>using namespace std;#define maxn 100010int pre[maxn];bool Set[maxn];int flag=1;int Find(int x){    while(pre[x]!=x)        x=pre[x];    return x;}void join(int x,int y){    int a=Find(x);    int b=Find(y);    if(a!=b)pre[a]=b;    else flag=0;//模板有这一步，习惯上带上}int main(){    //freopen("C.txt","r",stdin);    int n;    cin>>n;    while(n--)    {        int N,M;        cin>>N>>M;        for(int i=1;i<=N;i++)        {            pre[i]=i;            Set[i]=0;        }        int a,b,root_num=0;        while(M--)        {            cin>>a>>b;            if(a<b)join(a,b);  //习惯上把大的设为父亲            else join(b,a);        }        for(int i=1;i<=N;i++)        {            if(pre[i]==i)root_num++;//计算根的数目        }        cout<<root_num<<endl;    }    return 0;}

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