【HDU-1213】How Many Tables

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How Many Tables

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

Total Submission(s): 32574    Accepted Submission(s): 16254


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

//         题意： 分配桌子的问题,如果A认识B,B认识C,则ABC在一个桌子，D认识E,DE在另一个;

//        并查集： 认识的合并,看有几个根节点即安排几个桌子; 

//代码如下： 

#include <cstdio>
const int maxn = 1e6 + 10;
int par[maxn];
int ans = 0;
void init (int n) //父亲节点赋值为与其本身相同的值
{
for (int i = 1; i <= n ;i++)
par[i] = i;
}
int find(int x) //查找父亲节点
{
int p = x;
while (x != par[x])
{
x = par[x];
}
while (p != x)
{
int j = par[x];
par[x] = x;
p = j;
}
return x;
}
void unite(int a,int b) //合并
{
int fa = find(a);
int fb = find(b);
if (fa != fb)
{
par[fa] = fb;
}
}
int main()
{
int t;
scanf ("%d",&t);
while (t--)
{
int n,m;
int a,b;
ans = 0;
scanf ("%d%d",&n,&m);
init(n);
for (int i = 0 ; i < m ; i++)
{
scanf ("%d%d",&a,&b);
unite(a,b);
}
for (int i = 1 ; i <= n ; i++)
{
if (par[i] == i) //有几个根节点,就是答案。
ans++;
}
printf ("%d\n",ans);
}
return 0;
}