并查集

一、思想

通俗的说就是将所有关系的都连接起来，比方说A和B是朋友，B和C是朋友，那么A和C也是朋友，就算是中间隔着100个朋友这样的也算是朋友。

具体实现：

通过两个函数实现。

一是查找父节点函数，此函数可以使用递归调用来压缩路径。

二是合并函数，将两个没有连接的点连接起来。

二、例题 HDU-1213 How Many Tables？

题目地址：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

题目大意是求出有几个独立的集合。

AC代码：

#include<iostream>
#include<cstdio>
#include<string.h>

using namespace std;
int per[1000+10];
int root[1000+10];

int xfind( int a)
{
    if( a == per[a])
       return a;
    per[a] = xfind(per[a]);
    return per[a];
}

void mix( int a, int b)
{
    int f = xfind(a), l = xfind(b);
    if( f != l)
        per[l] = f;
}

int main()
{
    int t;
    int n, k;
    int a, b;

    cin >> t;
    while(t--)
    {
        int ans = 0;
        cin >> n >> k;
        for( int i = 1; i <= n; i++)
            per[i] = i;
        for( int i = 0; i < k; i++)
        {
            cin >> a >> b;
            mix( a, b);
        }
        memset( root, false, sizeof(root));
        for( int i = 1; i <= n; i++)
            root[xfind(i)] = 1;
        for( int i = 1; i <= n; i++)
            if(root[i])
                ans++;
        cout << ans << endl;
        getchar();
    }
    return 0;
}

带权并查集

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