寒假私训 —— 并查集 E - How Many Tables-优快云博客

本文链接：https://blog.youkuaiyun.com/atnanajiang/article/details/104123908

博客介绍了如何解决编程竞赛中的一道题目——E - How Many Tables。通过输入的朋友关系判断最少需要多少张桌子，使得每个朋友群体都能坐在一张桌子上。解题关键在于使用并查集来检查图的连通性，找出连通组件的数量。给出的解题思路和代码可帮助读者理解并查集的应用。

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E - How Many Tables

题目要求

 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

Sample Output

2
4

解题思路

该题和第一题A题思路差不多，检查连通性的，看有几个根节点，以判断有几个树，代码如下：

代码

#include<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define maxn 1000100
using namespace std;

 int f[maxn];

 int find(int x){
     if(x==f[x]){
         return x;
     }
     f[x]=find(f[x]);
     return f[x];
 }

 int main(){
     int t,i;
     scanf("%d", &t);
     for(i=1;i<=t;i++){
         int n,m;
         scanf("%d%d",&n,&m);
         for(int i=1; i<=n; i++){
             f[i] = i;
         }
         int sum=0;
         while(m--){
             int x, y;
             scanf("%d%d",&x,&y);
             int f1=find(x);
             int f2=find(y);
             if(f1!=f2)
                 f[f2]=f1;
         }

         for(int j=1;j<=n;j++){
               int a=find(j);
               if(a==j)
                              sum++;
         }
          printf("%d\n",sum);
     }
     return 0;
 }