九度OJ-1445：How Many Tables

　　用并查集求连通分量数。与上题完全一样，不赘述。

题目描述：

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.

输入：

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.

输出：

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

样例输入：

25 31 22 34 55 12 5

样例输出：

24

#include <iostream> #define MAXSIZE 1000using namespace std;struct UFS{//UnionFindSet ´Ó0¿ªÊ¼ int elem[MAXSIZE];int size;void initiate(int size){this->size=size;for (int i=0;i<size;i++)elem[i]=-1;}int findRoot(int x){//reduce deep if (elem[x]==-1) return x;else return elem[x]=findRoot(elem[x]);}void unionSet(int x,int y){int xroot=findRoot(x),yroot=findRoot(y);if (xroot!=yroot)//if x and y are from different set elem[xroot]=yroot; //merge xset into yset }};int main(){int t,n,m;int a,b;int count;UFS ufs;cin>>t;for (int time=0;time<t;time++){cin>>n>>m;//initiateufs.initiate(n);count=0;//input&&unionfor (int i=0;i<m;i++){cin>>a>>b;a-=1;b-=1;ufs.unionSet(a,b);}//countfor (int i=0;i<ufs.size;i++){if (ufs.elem[i]==-1)count++;} cout<<count<<endl;}return true;}