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
解题思路
题意说的是有n个朋友来吃晚餐，但是这些朋友中有某些人是互相认识的
这些互相认识的可以安排到一张桌子上
互相不认识就每人需要多拿出一张桌子
要我们求至少需要多少张桌子可以解决
我们先把互相认识的人unite起来，然后find一下是否属于同一个集合就可以解决问题了
AC代码
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 1010;
int Fa[maxn];
void init(int n)
{
for(int i = 1 ; i <= n ; i++)
{
Fa[i] = i;
}
}
int Find(int x)
{
return Fa[x]==x?x:Fa[x]=Find(Fa[x]);
}
void unite(int a, int b)
{
Fa[Find(a)] = Find(b);
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,m;
scanf("%d %d",&n,&m);
init(n);
int table = 0;
for(int i = 0 ; i < m ; i++)
{
int a,b;
scanf("%d %d",&a,&b);
unite(a,b);//合并a b
}
for(int i = 1 ; i <= n ; i++)
{
//合并完成后
//有关系的Fa[] = 较大的Fa[]
//他们共享较大的Fa[]
//如果等于i 既说明走到了较大的Fa[] 也说明走到了两人没有关系的
//故table++;
if(Fa[i] == i)//不同集合
{
table++;
}
}
printf("%d\n",table);
}
return 0;
}