POJ-The Suspects (disjoint-set template title)

加油。
To be better.

Topic links:
https://vjudge.net/problem/POJ-1611

Topic effect:
to give you a n and m, n represents the total number of students, m represents the number of groups, the next m lines, each input a number of student representative bodies in t, t enter the next number represents students in this group. Note that once the community might have a SARS virus infection, then the whole group on SARS virus are likely infected. 0 is the default number of possible SARS virus infection.

Solution:
This is actually a disjoint-set template title, but first start I stuck in how to make all of those infected index points to 0, in fact, not be used as such, requires only conventional combined collections, and finally from 1 to traverse n 0 and about who is an ancestor that Find (x) == Find (0 );

AC Code:

#include<iostream>
using namespace std;
struct node
{//我这里用的是结构题，方便路径压缩
    int x,pre,rak;
} f[30011];
int n,m;

void Init()
{//初始化
    for(int i=0; i<=n; i++)
    {
        f[i].pre=i;
        f[i].rak=1;
    }
}

int Find(int x)
{//查找函数
    if(f[x].pre==x)
        return x;
    else
        return Find(f[x].pre);
}

void Union(int x,int y)
{//合并（路径压缩优化）
    int rx=Find(f[x].pre);
    int ry=Find(f[y].pre);
    if(rx!=ry)
    {
        if(f[rx].rak<f[ry].rak)
            f[rx].pre=f[ry].pre;
        else
        {
            f[ry].pre=f[rx].pre;
            if(f[rx].rak==f[ry].rak)
                f[rx].rak++;
        }
    }
}

int Query()
{//查询一共有多少个可能的感染者
    int t=Find(f[0].pre);
    int ans=1;
    for(int i=1;i<=n;i++)
    {
        if(t==Find(f[i].pre))
            ans++;
    }
    return ans;
}
int main()
{
    int t,a,b;
    while(cin>>n>>m)
    {
        if(n==0)
            break;
        if(m==0)
            cout<<1<<endl;
        else
        {
            Init();
            for(int i=1; i<=m; i++)
            {
                cin>>t;
                for(int j=1; j<=t; j++)
                {
                    if(j==1)
                        cin>>a;//第一个数单独输入，方便后续合并
                    else
                    {
                        cin>>b;
                            Union(a,b);
                    }
                }
            }
            cout<<Query()<<endl;
        }
    }
    return 0;
}