并查集练习题

练习1

ZJU1789                                     The Suspects

Severe acute respiratory syndrome (SARS), an atypical pneumonia of unknown aetiology, was recognized as a global threat in mid-March 2003. To minimize transmission to others, the best strategy is to separate the suspects from others.
In the Not-Spreading-Your-Sickness University (NSYSU), there are many student groups. Students in the same group intercommunicate with each other frequently, and a student may join several groups. To prevent the possible transmissions of SARS, the NSYSU collects the member lists of all student groups, and makes the following rule in their standard operation procedure (SOP).
Once a member in a group is a suspect, all members in the group are suspects.
However, they find that it is not easy to identify all the suspects when a student is recognized as a suspect. Your job is to write a program which finds all the suspects.

Input

The input file contains several cases. Each test case begins with two integers n and m in a line, where n is the number of students, and m is the number of groups. You may assume that 0 < n <= 30000 and 0 <= m <= 500. Every student is numbered by a unique integer between 0 and n−1, and initially student 0 is recognized as a suspect in all the cases. This line is followed by m member lists of the groups, one line per group. Each line begins with an integer k by itself representing the number of members in the group. Following the number of members, there are k integers representing the students in this group. All the integers in a line are separated by at least one space.
A case with n = 0 and m = 0 indicates the end of the input, and need not be processed.

Output

For each case, output the number of suspects in one line.

Sample Input

100 4
2 1 2
5 10 13 11 12 14
2 0 1
2 99 2
200 2
1 5
5 1 2 3 4 5
1 0
0 0

Sample Output

4
1
1

#include <iostream>
#include <stdio.h>
#define N 30005

using namespace std;
int father[N];

void init(int n)
{
    int i;
    for(i = 0; i < n; i++)
    {
        father[i] = i;
    }
    return;
}

int getfather(int v)
{

    int geng = v;
    int temp;
    while(geng != father[geng])
    {
        geng = father[geng];
    }
    while(v != geng)
    {
        temp = father[v];
        father[v] = geng;
        v = temp;
    }
    return geng;

}

void Union(const int& i,const int& j)
{
    int x = getfather(i);
    int y = getfather(j);
    if(x == 0)
    {
        father[y] = 0;
    }
    else if(y == 0)
    {
        father[x] = 0;
    }
    else
        father[y] = x;


}
int main()
{
    int n, m;
    int k, a, b;
    while(~scanf("%d%d", &n, &m) && n + m)
    {
        int cot = 0;
        init(n);
        for(int i = 0; i < m; i++)
        {
            cin >> k;
            cin >> a;
            for(int j = 0; j < k - 1; j++)
            {
                cin >> b;
                Union(a, b);
            }
        }
        for(int i = 0; i < n; i++)
        {
            if(getfather(i) == 0)
            {
                cot++;
            }
        }
        cout << cot << endl;
    }
    return 0;
}

练习2

POJ2524                                    Ubiquitous Religions

There are so many different religions in the world today that it is difficult to keep track of them all. You are interested in finding out how many different religions students in your university believe in.

You know that there are n students in your university (0 < n <= 50000). It is infeasible for you to ask every student their religious beliefs. Furthermore, many students are not comfortable expressing their beliefs. One way to avoid these problems is to ask m (0 <= m <= n(n-1)/2) pairs of students and ask them whether they believe in the same religion (e.g. they may know if they both attend the same church). From this data, you may not know what each person believes in, but you can get an idea of the upper bound of how many different religions can be possibly represented on campus. You may assume that each student subscribes to at most one religion.

Input

The input consists of a number of cases. Each case starts with a line specifying the integers n and m. The next m lines each consists of two integers i and j, specifying that students i and j believe in the same religion. The students are numbered 1 to n. The end of input is specified by a line in which n = m = 0.

Output

For each test case, print on a single line the case number (starting with 1) followed by the maximum number of different religions that the students in the university believe in.

Sample Input

10 9
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
10 4
2 3
4 5
4 8
5 8
0 0

Sample Output

Case 1: 1
Case 2: 7

Hint

Huge input, scanf is recommended.

                          

#include <iostream>
#include <stdio.h>
#define N 50005

using namespace std;
int father[N];
int num[N];
int cot;
static int cas = 0;

void init(int n)
{
    int i;
    for(i = 1; i <= n; i++)
    {
        father[i] = i;
        num[i] = 1;
    }
    return;
}

int getfather(int v)
{
    int geng = v;
    int temp;
    while(geng != father[geng])
    {
        geng = father[geng];
    }
    while(v != geng)
    {
        temp = father[v];
        father[v] = geng;
        v = temp;
    }
    return geng;
}

void Union(int a, int b)
{
    int x = getfather(a);
    int y = getfather(b);
    if(x == y)
    {
        return;
    }
    else
    {
        cot -= 1;
        if(num[x] > num[y])
        {
            father[y] = x;
            num[x] += num[y];
        }
        else
        {
            father[x] = y;
            num[y] += num[x];
        }

    }
}

bool judge(int a, int b)
{
    if(getfather(a) == getfather(b))
        return true;
    else
        return false;

}

int main()
{
    int n;
    long long m;
    int a, b;
    while(~scanf("%d%I64d", &n, &m) && (n + m))
    {
        init(n);
        cot = n;
        for(int i = 1; i <= m; i++)
        {
            scanf("%d%d", &a, &b);
            Union(a, b);
        }
        printf("Case %d: %d\n", ++cas, cot);
    }
}