How Many Tables

xiaoxiao2021-02-28  45

 How Many Tables

 

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: n个人一起去饭店吃饭,相互认识或间接认识的人可以坐一桌,问最少需要几张桌子。 思路:把相互认识的人合并,看最后剩下几个互不认识的团 即 fa[i]==i的个数 简单运用并查集模板, 代码:   #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; const int N=3000+20; int fa[N]; int find(int x) { if(fa[x]==x) return x; return find(fa[x]); } void join(int x,int y) { int fx,fy; fx=find(x),fy=find(y); if(fx!=fy) fa[fy]=fx; } int main() { int t; scanf("%d",&t); while(t--) { int n,m; scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) fa[i]=i; int x,y; for(int i=0;i<m;i++) { scanf("%d%d",&x,&y); join(x,y); } int cnt=0; for(int i=1;i<=n;i++) if(fa[i]==i) cnt++; printf("%d\n",cnt); } return 0; }

 

转载请注明原文地址: https://www.6miu.com/read-2300040.html

最新回复(0)