HDU1213How Many Tables(基础并查集)
HDU1213How Many Tables
Problem Description
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
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
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
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=20000;
int pre[maxn],height[maxn];
void init_set(int n){
for (int i = 1; i <= n; i++){
pre[i]=i;
}
memset(height,0,sizeof(height));
} int find_set(int x){
return x==pre[x]?x:pre[x]=find_set(pre[x]);
}
void union_set(int x,int y){
x= find_set(x);
y= find_set(y);
if(x==y)return;
if(height[x]==height[y]){
height[x]=height[x]+1;
pre[y]=x;
}else{
if(height[x]<height[y]) pre[x]=y;
else{
pre[y]=x;
}
}
} int main(){
int T;
cin>>T;
int n,m,a,b;
while (T--){
cin>>n>>m;
init_set(n);
int sum=0;
while (m--){
cin>>a>>b;
union_set(a,b);
}
for (int i = 1; i <=n; i++)
{
if(i==pre[i])sum++;
}
cout<<sum<<endl;
}
}
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