poj 1611:The Suspects（并查集，经典题）

The Suspects

Time Limit: 1000MS		Memory Limit: 20000K
Total Submissions: 21472		Accepted: 10393

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

Source

Asia Kaohsiung 2003

---

 

　　并查集，经典题。

　　并查集模板（来自北大课件）：

int ufs[MAXN];    //并查集

void Init(int n)    //初始化
{
    int i;
    for(i=0;i<n;i++){
        ufs[i] = i;
    }
}

int GetRoot(int a)    //获得a的根节点。路径压缩
{
    if(ufs[a]!=a){    //没找到根节点
        ufs[a] = GetRoot(ufs[a]);
    }
    return ufs[a];
}

void Merge(int a,int b)    //合并a和b的集合
{
    ufs[GetRoot(b)] = GetRoot(a);
}

bool Query(int a,int b)    //查询a和b是否在同一集合
{
    return GetRoot(a)==GetRoot(b);
}

 

　　题意：

　　

 

　　思路：

　　很经典的并查集的题目，找一个sum[]数组记录每一个以当前下标为根节点的集合的个体数目，最后输出0号的根节点对应的sum值，就是0号学生所在团体的人数。

 

　　代码：

#include <iostream>
#include <stdio.h>
using namespace std;
#define MAXN 30010 
int sum[MAXN];    //集合总数
int ufs[MAXN];    //并查集

void Init(int n)    //初始化
{
    int i;
    for(i=0;i<n;i++){
        ufs[i] = i;
        sum[i] = 1;
    }
}

int GetRoot(int a)    //获得a的根节点。路径压缩
{
    if(ufs[a]!=a){    //没找到根节点
        ufs[a] = GetRoot(ufs[a]);
    }
    return ufs[a];
}

void Merge(int a,int b)    //合并a和b的集合
{
    int x = GetRoot(a);
    int y = GetRoot(b);
    if(x!=y){
        ufs[y] = x;
        sum[x] += sum[y];
    }
}

int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF){
        if(n==0 && m==0) break;
        Init(n);    //初始化并查集
        while(m--){    //读入m行
            int t,one,two;
            scanf("%d",&t);    //每一行有t个数需要输入
            scanf("%d",&one);
            t--;
            while(t--){
                scanf("%d",&two);
                Merge(one,two);    //合并集合
            }
        }
        printf("%d\n",sum[GetRoot(0)]);
    }
    return 0;
}

 

Freecode : www.cnblogs.com/yym2013