【并查集】How Many Tables

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原文链接：http://www.cnblogs.com/Wurq/articles/4553545.html

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#java

本文介绍如何使用并查集算法解决社交派对中朋友间相互认识的桌次分配问题，通过输入的朋友数量和相互认识的关系，输出所需最少的桌子数量。

How Many Tables

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17271 Accepted Submission(s): 8456

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

Author

Ignatius.L

题目大意：寻找有多少个独立的点。。。

并查集直接做，判断没有重复的边M，然后独立的点=N-M就是答案、

 1 #include <iostream>
 2 #include <string.h>
 3 #include <stdio.h>
 4 #include <map>
 5 using namespace std;
 6 int ID[10086];
 7 void Cread(int N)
 8 {
 9     for(int i=0;i<=N;i++)ID[i]=i;
10 }
11 int Find(int x)
12 {
13     int tmp;
14     if(ID[x]!=x)tmp=Find(ID[x]);
15     else tmp=x;
16     ID[x]=tmp;
17     return tmp;
18 }
19 int main()
20 {
21     int N,M,i,j,k,T;
22     scanf("%d",&T);
23     while(T--)
24     {
25         scanf("%d%d",&N,&M);
26         Cread(N);
27         int a,b;
28         int sign=0;
29         for(i=0;i<M;i++)
30         {
31             scanf("%d%d",&a,&b);
32             a=Find(a);
33             b=Find(b);
34             if(a!=b)
35             {
36                 sign++;
37                 ID[a]=b;
38             }
39         }
40         printf("%d\n",N-sign);
41     }
42     return 0;
43 }