HDU1213

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.

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

Sample Output

2
4

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int parent[1010],ans;
 4 int i,j,t,m,n,a,b,f1,f2;
 5 int Find(int x)
 6 {
 7     if(parent[x]!=x)return x=Find(parent[x]);
 8     return x;
 9 }
10 /*
11 void union_(int x,int y)
12 {
13     if((x=Find(x)==(y=Find(y))))
14         return;
15     if(parent[x]<parent[y])
16     {
17         parent[x]+=parent[y];
18         parent[y]=x;
19     }
20     else
21     {
22         parent[y]+=parent[x];
23         parent[x]=y;
24     }
25 }
26 */
27 int main()
28 {
29    cin>>t;
30    for(i=1;i<=t;i++)
31    {
32        ans=0;
33        cin>>m>>n;
34        for(j=1;j<=m;j++)
35         parent[j]=j;
36        for(j=1;j<=n;j++)
37        {
38            cin>>a>>b;
39            f1=Find(a);
40            f2=Find(b);
41            parent[f1]=parent[f2];
42        }
43        for(j=1;j<=m;j++)
44         if(parent[j]==j)
45         ans++;
46        cout<<ans<<endl;
47    }
48     return 0;
49 }

思路：

思路简单就是查找有多少个不同的联通分量。

注意点：

一开始做这个题还多写了个void union_(int x,int y)，被MLE了。后来想了下原因，可能是这个函数在执行的时候进行了多次Find(x),而这个函数是通过递归来实现的，递归的实现要占用内存，所以内存就上去了。后来用一个赋值语句代替了这个函数parent[f1]=parent[f2]; 。因为本题只需要求出不同连通分量的个数，不需要知道一个联通分量里元素的个数，所以没有必要再写原来那个函数了。