how many tables(并查集)

How Many Tables

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 19  Solved: 12
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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

HINT

#include<stdio.h>#include<iostream>#include<string.h>#include<algorithm>using namespace std;#define MAX 100001int pre[1001];int find(int a){    int r=a;    while(r!=pre[r])        r=pre[r];        int i=a,j;        while(i!=r)        {            j=pre[i];            pre[i]=r;            i=j;        }    return r;}void join(int a,int b){    int x=find(a);    int y=find(b);    if(x!=y)        pre[x]=y;}void slove(){    for(int i=0;i<=1001;i++)        pre[i]=i;}int main(){    int t;    cin>>t;    while(t--)    {        slove();        int m,n;         int x,y;        cin>>m>>n;        while(n--)        {            cin>>x>>y;            join(x,y);        }        int count;        int i;        for(count=0,i=1;i<=m;i++)            //假设现在一人一桌，现在如果谁的pre[i]!=i;就说明i到别人的桌上去了            //现在只要看谁还在自己的桌上就说明还需要几桌        {//            cout<<pre[i]<<" ";            if(pre[i]==i)                count++;        }        cout<<count<<endl;//        cout<<endl;    }    return 0;}