# (并查集 注意别再犯傻逼的错了) HDU 1213 How Many Tables

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

emmmmm，我交了10多次才发现我在一个小地方上犯了傻逼的错。。。。
C++代码：
#include<iostream>
#include<stdio.h>
#include<cstring>
using namespace std;
const int maxn = 1010;
int father[maxn],vis[maxn];
int Find(int x){
while(x != father[x]){
father[x] = father[father[x]];
x = father[x];
}
return x;

}
void Union(int a,int b){
int ax = Find(a);
int bx = Find(b);
if(ax != bx){
father[ax] = bx;
}
//    if(a != b)
//        father[a] = b; //这个错的太傻逼了吧
}
int main(){
int T;
scanf("%d",&T);
int a,b;
while(T--){
int n,m;
scanf("%d%d",&n,&m);
memset(vis,0,sizeof(vis));
for(int i = 1; i <= n; i++)
father[i] = i;
for(int i = 0; i < m; i++){
scanf("%d%d",&a,&b);
Union(a,b);
}
int sum = 0;
for(int i = 1; i <= n; i++){
if(father[i] == i)
sum++;
}
cout<<sum<<endl;
}
return 0;
}

posted @ 2019-05-22 19:52  PJCK  阅读(99)  评论(0编辑  收藏  举报