acm集训训练赛（二）D题【并查集】

一、题目

Description

There is a town with N citizens. It is known that some pairs of people are friends. According to the famous saying that The friends of my friends are my friends, too it follows that if A and B are friends and B and C are friends then A and C are friends, too.

Your task is to count how many people there are in the largest group of friends.

Input

Input consists of several datasets. The first line of the input consists of a line with the number of test cases to follow. The first line of each dataset contains tho numbers N and M, where N is the number of town's citizens (1≤N≤30000) and M is the number of pairs of people (0≤M≤500000), which are known to be friends. Each of the following M lines consists of two integers A and B (1≤A≤N, 1≤B≤N, A≠B) which describe that A and B are friends. There could be repetitions among the given pairs.

Output

The output for each test case should contain one number denoting how many people there are in the largest group of friends.

Sample Input

Sample Output

3 2

1 2

2 3

10 12

1 2

3 1

3 4

5 4

3 5

4 6

5 2

2 1

7 10

1 2

9 10

8 9

二、题目源程序

#include <cstdio>
#include <iostream>
#include <cmath>using namespace std;int parent[30005];
int num[30005];
int n, m;
int find (int x) {return parent[x] == x ? x : find(parent[x]);
}
void init () {int i;for (i = 0; i < n; i ++)parent[i] = i;for (i = 0; i < n; i ++)num[i] = 0;
}
void Union(int a, int b) {int fa = find(a);int fb = find(b);if(fa != fb) {parent[fa] = fb;}
}
int main () {int cases;int a, b;int i;scanf("%d", &cases);while (cases --) {scanf("%d%d", &n, &m);init ();for (i = 0; i < m; i ++) {scanf("%d%d", &a, &b);Union(a, b);}for (i = 0; i < n; i ++) {int x = find(i);num[x] ++;}int max = 0;for (i = 0; i < n; i ++) {if(num[i] > max)max = num[i];}printf("%d\n", max);}return 0;
}

三、解题思路

一个简单的运用并查集的知识的题目。

练习基本的并查集的运用可以用这道题。

转载于:https://www.cnblogs.com/fightfor/p/3871547.html