HDOJ1213 How Many Tables 并查集裸题

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本文介绍了一个经典的并查集问题，通过使用并查集算法解决如何计算参加生日派对的朋友最少需要多少张桌子的问题。算法首先初始化每个朋友为单独的一组，然后根据朋友之间的熟识关系进行合并，最后统计剩余的组数即为所需最少桌子数量。

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

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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;
}