POJ 1611 The Suspects (并查集+数组记录子孙个数 )

The Suspects

Time Limit: 1000MS		Memory Limit: 20000K
Total Submissions: 24134		Accepted: 11787

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

 

题目及算法分析：有n个人，编号0---n-1，m个组，输入这m组每组人的编号，每个组都是一个小集体。

     假设0号人是嫌疑人，所有跟嫌疑人在一个组的都是嫌疑人。注意：如果1和0在某个组，并且1和2, 3, 4等在一个组，这些人也都是嫌疑人！

采用并查集算法进行合并计算算法，需要开辟一个数组进行0号，进行每个节点的子孙数目的统计保存。

代码：

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <ctype.h>
#include <iostream>
#include <string>
#include <iomanip>
#include <algorithm>
#define N 30000

using namespace std;
int n, m;
int cnt[N];
int fa[N];

//0 < n <= 30000 and 0 <= m <=500
void init()
{
    for(int i=0; i<N; i++)
    {
        fa[i]=i;
        cnt[i]=1;
    }
}

int findset(int x)
{
    return fa[x]!=x? fa[x]=findset(fa[x]):x;
}

void union_set(int x, int y)
{
    int xx=findset(x);
    int yy=findset(y);
    if(xx==yy) return; //说明两元素本来就属于同一个集合 返回
    else if(xx<yy) //如果x的根节点比y的根节点 小
    {
        fa[yy]=xx;
        cnt[xx]=cnt[xx]+cnt[yy];
    }
    else if(xx>yy)
    {
        fa[xx]=yy;
        cnt[yy]=cnt[yy]+cnt[xx];
    }
}

int main()
{
    int dd, a, b;
    while(scanf("%d %d", &n, &m)!=EOF)
    {
        if(n==0 && m==0) break;
        init();
        for(int i=0; i<m; i++)
        {
            scanf("%d", &dd);
            scanf("%d", &a);
            for(int j=0; j<dd-1; j++)
            {
                scanf("%d", &b);
                union_set(a, b);
                a=b;
            }
        }
        printf("%d\n", cnt[0] );
    }
    return 0;
}