HDOJ1213 How Many Tables 并查集裸题

关注

原创最新推荐文章于 2026-07-09 11:12:32 发布 298 阅读 AI 写同款· GEO 优化 ›

并查集(Union-Find) 应用举例 --- 基础篇本文是作为上一篇文章《并查集算法原理和改进》的后续，焦点主要集中在一些并查集的应用上。材料主要是取自POJ，HDOJ上的一些算法练习题。首先还是回顾和总结一下关于并查集的几个关键点：以树作为节点的组织结构，结构的形态很是否采取优化策略有很大关系，未进行优化的树结构可能会是“畸形”树(严重不平衡，头重脚轻，退化成链表等)，按尺寸(正规说法叫做秩，后文全部用秩来表示)进行平阅读详情

How Many Tables

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 33675 Accepted Submission(s): 16840

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

Sample Output

2
4

Author

Ignatius.L

Source

杭电ACM省赛集训队选拔赛之热身赛

Recommend

Eddy | We have carefully selected several similar problems for you: 1856 1325 1198 1879 1875

裸题，程序就是标准的模板

#include <iostream>
#include <cmath>
#include <ctime>
using namespace std;

const int maxn = 1005;
int n,m,i,j,k,tot;
int root[maxn];

void init(){
   for (i=1; i<=n; i++) root[i] = i;
   tot = 0;
}

int Find(int x){
   int r = x;
   while (r!=root[r]) r = root[r];
   int i = x, k;
   while (root[i] != r) {
      k = root[i];
      root[i] = r;
      i = k;
   }
   return r;
}

void Merge(int x, int y){
   int fx = Find(x), fy = Find(y);
   if (fx != fy) root[fx] = fy;
}

int main(){
    std::ios::sync_with_stdio(false);
    int t;
    cin >> t;
    while (t--) {
        cin >> n >> m;
        init();
        for (k=1; k<=m; k++) {
            cin >> i >> j;
            Merge(i,j);
        }
        for (i=1; i<=n; i++) if (root[i]==i) tot++;
        cout << tot << endl;
    }
    return 0;
}

收录于