POJ1611(The Suspects)--简单并查集

题目在这里

关于SARS病毒传染的问题。在同一个组的学生是接触很近的，后面也会有新的同学的加入。其中有一位同学感染SARS，那么该组的所有同学得了SARS。要计算出有多少位学生感染SARS了。编号为0的同学是得了SARS的。

直接用并查集解决水掉

#include<iostream>
#include<stdio.h> 
#include<cstring>
#include<cmath>
#include<vector>
#include<stack>
#include<map>
#include<set>
#include<list>
#include<queue>
#include<string>
#include<algorithm>
#include<iomanip>
using namespace std;
#define MAX 100

struct node
{
    int no;//编号
    int rank;
    int parent;
    int total;
};

class DisJoinSet
{
    protected:
        int n;
        node * tree;
    public:
        DisJoinSet(int n );
        ~DisJoinSet();
        void Init();
        void Union(int x,int y);
        int Find(int x);
        int GetAswer(int x);
};

DisJoinSet ::DisJoinSet(int  n)//初始化操作
{
    this->n = n;
    tree = new node[n];
    for(int i = 0 ; i < n; i++)
    {
        tree[i].no = i;
        tree[i].parent = i;
        tree[i].total = 1;
        tree[i].rank = 0;
    }
}
DisJoinSet::~DisJoinSet()
{
    delete[] tree;
}

void DisJoinSet :: Init()
{
}
int DisJoinSet::Find(int x)
{
    int temp = tree[x].parent;//temp 为x的父亲结点
    if( x != tree[x].parent)
    {
        tree[x].parent = Find(tree[x].parent);//路径压缩
        return tree[x].parent;
    }
    else
    {
        return x;
    }
}

void DisJoinSet ::Union(int x,int y)
{
    int rootx = Find(x);
    int rooty = Find(y);
    if(rootx == rooty)
    {
        return ;
    }
    else//并查集基本操作
    {
        if(tree[rootx].rank < tree[rooty].rank)
        {
            tree[rootx].parent = rooty;
            tree[rooty].total += tree[rootx].total;
        }
        else
        {
            tree[rooty].parent = rootx;
            tree[rootx].total += tree[rooty].total;
            if(tree[rootx].rank == tree[rooty].rank)
            {
                tree[rootx].rank++;
            }
        }
    }
}

int DisJoinSet::GetAswer(int x)//返回xtotal
{
    int t = Find(x);
    return tree[t].total;
}

int main()
{
    int n;
    int m;
    while(cin>>n>>m && n!=0 && m>= 0)
    {
        DisJoinSet dis(n);
        int per1;
        int per2;
        for(int i = 0; i< m;i++)
        {
            int num = 0;
            cin>>num;
            cin>>per1;
            for(int i = 1 ;i < num; i++)
            {
                cin>>per2;
                dis.Union(per1,per2);
            }
        }
        cout<<dis.GetAswer(0)<<endl;
    }
    return 0;
}