PKU POJ 1611解题报告(并查集)

文章作者：yx_th000 文章来源：Cherish_yimi (http://www.cnblogs.com/cherish_yimi/) 转载请注明，谢谢合作。

并查集学习--并查集详解

The Suspects

Time Limit: 1000MS		Memory Limit: 20000K
Total Submissions: 5572		Accepted: 2660

Description

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

我的思路：
典型的并查集，最初各自为集，然后每个group进行合并，等到所有的group合并完，题目也就解决了，因为在合并的时候，如果哪两个group中有重合的元素，则那个后来的group会由于按秩合并的原则自动合并到
先有的集合当中，奥妙便在其中。下面是代码：

 1#include<iostream>
 2using namespace std;
 3
 4int n, m, i, j;
 5int father[30005], num[30005];
 6
 7void makeSet(int n)
 8{
 9    for(i = 0; i < n; i++)
    {
        father[i] = i; //使用本身做根
        num[i] = 1;
    }
}
int findSet(int x)
{
    if(father[x] != x) //合并后的树的根是不变的
    {    
        father[x] = findSet(father[x]);
    }
    return father[x]; 
}

void Union(int a, int b)
{
    int x = findSet(a);
    int y = findSet(b);
    if(x == y)
    {
        return;
    }
    if(num[x] <= num[y])
    {
        father[x] = y;
        num[y] += num[x];
    }
    else 
    {
        father[y] = x;
        num[x] += num[y];
    }
}

int main()
{
    while(scanf("%d %d", &n, &m)!=EOF && n != 0)
    {
        makeSet(n);
        for(i = 0; i < m; i++)
        {
            int count, first, b;
            scanf("%d %d",&count, &first);
            for(j = 1; j < count; j++)
            {
                scanf("%d",&b);
                Union(first,b);
            }
        }
        printf("%d\n",num[findSet(0)]);
    }
    return 0;
}

另外，上面并查集的根我是采用数字本身的，然后路径压缩寻找父亲节点是采用递归的，下面贴出，采用-1做根和使用非递归实现的部分代码：

 1void makeSet(int n)
 2{
 3    for(i = 0; i < n; i++)
 4    {
 5        father[i] = -1;
 6        num[i] = 1;
 7    }
 8}
 9//非递归实现
10int findSet(int x)
11{
12    while(father[x] != -1)   //根为-1
13    {
14        x = father[x];
15    }
16    return x;
17}
18