POJ-1213 How Many Tables( 并查集 )

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1213

Problem 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

2

5 3

1 2

2 3

4 5

 

5 1

2 5

 

Sample Output

2 4

 

水题一个，没什么可说的。不过这题意外的让我发现我以前的并查集学错了，真是意外收获，看来偶尔做点水题也是极好的。

 

 #include<iostream>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<iomanip>
 #include<map>

 using namespace std;

 int fa[];
 bool flag[];

 int findf( int x ){
     int root = x, pos = x;
     while( root != fa[root] ){
         root = fa[root];
     }
     while( pos != root ){
         x = fa[pos];
         fa[pos] = root;
         pos = x;
     }
     return root;
 }

 int main(){
     ios::sync_with_stdio( false );

     int t, n, m, a, b, ans;
     cin >> t;
     while( t-- ){
         cin >> n >> m;
         for( int i = ; i <= n; i++ ){
             fa[i] = i;
         }
         memset( flag, false, sizeof( flag ) );
         ans = ;

         while( m-- ){
             cin >> a >> b;
             if( findf( a ) != findf( b ) ){
                 fa[findf( a )] = b;
             }
         }

         for( int i = ; i <= n; i++ ){
             int temp = findf( i );
             if( !flag[temp] ){
                 ans++;
                 flag[temp] = true;
             }
         }

         cout << ans << endl;
     }

     return ;
 }