C - How Many Tables 并查集
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.
InputThe 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.
OutputFor 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 这个题做麻烦了,其实可以直接遍历pre数组,其中pre[i]==i的个数就是所有集合的个数!
#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<cstring>
#include<cmath>
#include<vector>
#include<set>
#include<iomanip>
#include<iostream>
using namespace std;
#define MAXN 30000
#define INF 0x3f3f3f3f
/*
并查集
*/
set<int> s;
int pre[MAXN],n,m;
int find(int x)
{
if(x==pre[x])
return x;
else
return pre[x] = find(pre[x]);
}
void mix(int x,int y)
{
int fx = find(x),fy = find(y);
if(fx!=fy)
pre[fy] = fx;
}
int main()
{
int t;
cin>>t;
while(t--)
{
int x,y;
cin>>n>>m;
for(int i=;i<=n;i++)
pre[i] = i;
s.clear();
for(int i=;i<m;i++)
{
cin>>x>>y;
mix(x,y);
}
int cnt = ;
for(int i=;i<=n;i++)
{
int fi = find(i);
if(!s.count(fi))
{
cnt++;
s.insert(fi);
}
}
cout<<cnt<<endl;
}
return ;
}
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