思路：
对我这种新手来说挺难想到的并查集吧，好歹看了题解能明白了
每一行和每一列都看成并查集中的一个点，所以一共有n+m个点，
可以看出，只要一个矩形只要三个角上的点有共同的祖先，那么第四个点的x和y的祖先肯定也是它（可以模拟一下），所以出现一个点就将这个点的x和y合并，最后循环n+m 看哪一行或者哪一边没有加进来，遇到一个让ans++，注意最后的结果是 ans-1
B. Chemical table
time limit per test
1 second
memory limit per test
512 megabytes
input
standard input
output
standard output
Innopolis University scientists continue to investigate the periodic table. There are n·m known elements and they form a periodic table: a rectangle with n rows and m columns. Each element can be described by its coordinates (r, c) (1 ≤ r ≤ n, 1 ≤ c ≤ m) in the table.
Recently scientists discovered that for every four different elements in this table that form a rectangle with sides parallel to the sides of the table, if they have samples of three of the four elements, they can produce a sample of the fourth element using nuclear fusion. So if we have elements in positions (r1, c1), (r1, c2), (r2, c1), where r1 ≠ r2 and c1 ≠ c2, then we can produce element (r2, c2).
Samples used in fusion are not wasted and can be used again in future fusions. Newly crafted elements also can be used in future fusions.
Innopolis University scientists already have samples of q elements. They want to obtain samples of all n·m elements. To achieve that, they will purchase some samples from other laboratories and then produce all remaining elements using an arbitrary number of nuclear fusions in some order. Help them to find the minimal number of elements they need to purchase.
Input
The first line contains three integers n, m, q (1 ≤ n, m ≤ 200 000; 0 ≤ q ≤ min(n·m, 200 000)), the chemical table dimensions and the number of elements scientists already have.
The following q lines contain two integers ri, ci (1 ≤ ri ≤ n, 1 ≤ ci ≤ m), each describes an element that scientists already have. All elements in the input are different.
Output
Print the minimal number of elements to be purchased.
Examples
input
Copy
2 2 3 1 2 2 2 2 1output
Copy
0input
Copy
1 5 3 1 3 1 1 1 5output
Copy
2input
Copy
4 3 6 1 2 1 3 2 2 2 3 3 1 3 3output
Copy
1Note
For each example you have a picture which illustrates it.
The first picture for each example describes the initial set of element samples available. Black crosses represent elements available in the lab initially.
The second picture describes how remaining samples can be obtained. Red dashed circles denote elements that should be purchased from other labs (the optimal solution should minimize the number of red circles). Blue dashed circles are elements that can be produced with nuclear fusion. They are numbered in order in which they can be produced.
Test 1
We can use nuclear fusion and get the element from three other samples, so we don't need to purchase anything.
Test 2
We cannot use any nuclear fusion at all as there is only one row, so we have to purchase all missing elements.
Test 3
There are several possible solutions. One of them is illustrated below.
Note that after purchasing one element marked as red it's still not possible to immidiately produce the middle element in the bottom row (marked as 4). So we produce the element in the left-top corner first (marked as 1), and then use it in future fusions.
AC代码：
#include<iostream>
#define MAX 400010
using namespace std;
int n,m,q;
int f[MAX+1];
void init()
{
for(int i=1;i<=MAX;i++)
{
f[i]=i;
}
}
int getf(int x)
{
if(f[x]==x) return x;
else return f[x]=getf(f[x]);
}
void merge(int a,int b)
{
int t1=getf(a);
int t2=getf(b);
if(t1!=t2)
{
f[t1]=t2;
}
}
int main()
{
int ans=0;
scanf("%d%d%d",&n,&m,&q);
init();
while(q--)
{
int a,b;
scanf("%d%d",&a,&b);
merge(a,b+n);
}
for(int i=1;i<=n+m;i++)
if(f[i]==i) ans++;
printf("%d\n",ans-1);
return 0;
}