Friends(并查集)

F - Friends
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Time Limit:3000MS
Memory Limit:0KB 64bit IO Format:%lld & %llu

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Status
Practice
UVA 10608

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 (on a line by itself) one number denoting how many people there are in the largest group of friends on a line by itself.

Sample Input

2 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 1 1 2 9 10 8 9

Sample Output

3 7

#include <iostream>

#include <cstdio>

#include <algorithm>

#include <cstring>

using namespace std;

const int MAXN = 30030;

struct Fri

{

int f, num;

} father[MAXN];

int ran[MAXN];

int fin(int v)

{

return father[v].f = father[v].f == v ? v : fin(father[v].f);

}

void merg(int x, int y)

{

int a = fin(x), b = fin(y);

if( a == b )

return;

if( ran[a] < ran[b])

{

father[a].f = b;

father[b].num += father[a].num;

}

else

{

father[b].f = a;

father[a].num += father[b].num;

if(ran[b] == ran[a])

{

++ran[a];

}

}

}

int main()

{

int t, x, y, a, b, ans = 0, n, m;

cin >> t;

while(t--)

{

ans = 0;

memset(father, 0, sizeof(father));

memset(ran, 0, sizeof(ran));

cin >> n >> m;

for( int i=0; i<n; i++ )

{

father[i].f = i;

father[i].num = 1;

}

for( int i = 0; i<m; i++ )

{

cin >> a >> b;

merg(a, b);

x = fin(a);

y = fin(b);

ans = max(ans, max(father[x].num, father[y].num));

}

cout << ans << endl;

}

return 0;

}

最后我才发现我是看错题了，citizens都会默了///