HDU 1213 How Many Tables【简单并查集】

How Many Tables

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 30094    Accepted Submission(s): 14878

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

25 31 22 34 55 12 5

 

Sample Output

24

 

Author

Ignatius.L

 

Source

杭电ACM省赛集训队选拔赛之热身赛

 

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考点：并查集的简单应用

题意：有n个客人，给定m组数据代表两个人认识。给客人们安排桌子，客人们必须与自己认识的人坐在一桌。规则是A认识B，B认识C，则A认识C。

题解：使用并查集将所有认识的人放到一组，求总组数。

代码：

#include <iostream>using namespace std;const int maxn = 1010;int f[maxn];int Find(int t){    return f[t] == t ? t : f[t] = Find(f[t]);}void Union(int a, int b){    int p1 = Find(a), p2 = Find(b);    if (p1 != p2)        f[p1] = p2;}int main(){    ios::sync_with_stdio(false);    cin.tie(0);    int t, m, n, a, b, cnt;    cin >> t;    while (t--)    {        cnt = 0;        cin >> n >> m;        for (int i = 1; i <= n; i++)            f[i] = i;        for (int i = 1; i <= m; i++)        {            cin >> a >> b;            Union(a, b);        }        for (int i = 1; i <= n; i++)            if(f[i] == i)                cnt++;        cout << cnt << endl;    }    return 0;}

（有个小地方写错了，结果导致memory limit exceeded，真是罪过！查了下原因：会导致无限递归然后造成超内存。）

The end.