HDU 1213 How Many Tables 第一道并查集的题。

How Many Tables

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 15083    Accepted Submission(s): 7361


[align=left]Problem Description[/align]
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.

 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
For each test case, just output how many tables Ignatius needs at least. Do NOT print any blanks.

 

[align=left]Sample Input[/align]

2
5 3
1 2
2 3
4 5

5 1
2 5

 

[align=left]Sample Output[/align]

2
4

 

[align=left]Author[/align]
Ignatius.L
 

[align=left]Source[/align]
杭电ACM省赛集训队选拔赛之热身赛
 

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代码不是很长，可是我理解了很久很久，也不知道自己的理解是不是对的。反正分享一下。。
5 3
1 2
2 3
4 5
 
2
 
这个案例代表有编号为1到5的五个人， 三代表有三对朋友，下面分别代表1和2是朋友，2和3是朋友，4和5是朋友。。 输出2代表两张桌子。因为1和2是朋友，2和3是朋友，默认1和3也是朋友。所以三个人只要一张桌子。4和5一张桌子就够了。题目的意思就是这个。给你一个编号n，下面列出m对人的关系，问你最少需要准备多少张桌子。一个多月前我就做过这道题，那时很天真，随便想了个方法，结果了WA了，还跑去讨论区问别人（丢脸）。
 



 
以前天真的代码。。
现在我附上我所理解的代码。不知道有没理解错的。
 
 

#include <stdio.h>
int p[1005];
int find(int x)     //这个的作用就是下面的查找。
{
if(x!=p[x])
p[x]=find(p[x]);
return p[x];
}
int hebing(int x,int y)  //这个的作用就是用来合并的。
{
return  p[x]=y;     //假设a=2,b=3，此时应该有p[2]=p[3]=3。即2和3到同一张桌子了。
}
int main()
{
int t,i,a,b;
scanf("%d",&t);
while(t--)
{
int n,m,ans=0;
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++)
p[i]=i;   //初始化它，让编号为一的值为1，编号为2的值为2.以此类推。
for(i=1;i<=m;i++)
{
scanf("%d%d",&a,&b);
a=find(a);
b=find(b);   //假设a=2,b=3,我认为经过这个查找之后p[2]就等于p[3]了。
if(a!=b)
hebing(a,b);  //合并为一个值。
}
for(i=1;i<=n;i++)
{
if(p[i]==i)   //经过M次合并之后，如果是朋友，或者间接朋友的，他们对应的值都为同一个。所以桌子就减少了。
ans++;     //如果值还是对应的，那么就不是朋友关系，增加一张桌子。
}
printf("%d\n",ans);//输出注意格式，只有一个空行。
}
return 0;
}