POJ--1611--The Suspects

The Suspects

Time Limit: 1000MS Memory Limit: 20000KTotal Submissions: 18738 Accepted: 9052

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 42 1 25 10 13 11 12 142 0 12 99 2200 21 55 1 2 3 4 51 00 0

Sample Output

411

Source

Asia Kaohsiung 2003

 

 

Source Code

import java.util.Scanner;public class Main {/** * @param args */static int[] father=new int[30001];static int[] b=new int[]{1,1};public static void main(String[] args) {// TODO Auto-generated method stubScanner input=new Scanner(System.in);while(true){int n=input.nextInt();int m=input.nextInt();if(n==0&&m==0)//情况1break;if(n!=0&&m==0){//情况2System.out.println(n);continue;}//情况3：先建树，再判断initbrr(n);//初始化树int max=0;while(m-->0){int k=input.nextInt();int[] arr=new int[k];for(int i=0;i<k;i++){arr[i]=input.nextInt();if(arr[i]>max)max=arr[i];}F(arr);//利用数组，每组进行建树}int count=0;//记录嫌疑数量findFather(0);int p=b[0];for(int i=1;i<=max;i++){findFather(i);int q=b[0];if(p==q)//判断i是否和0在同一棵树上；只要同一棵树上，就会被视为嫌疑count++;}System.out.println(count+1);//不要忘记没加0，i是从1开始的}}private static void initbrr(int n) {// TODO Auto-generated method stubfor(int i=1;i<father.length;i++)father[i]=i;}private static void F(int[] arr) {// TODO Auto-generated method stubfor(int i=1;i<arr.length;i++)FF(arr[0],arr[i]);//都和开头的数进行建树}private static void FF(int i, int j) {// TODO Auto-generated method stubfindFather(i);int gen1=b[0];int len1=b[1];findFather(j);int gen2=b[0];int len2=b[1];if(len1>=len2)father[gen2]=gen1;elsefather[gen1]=gen2;}private static void findFather(int i) {// TODO Auto-generated method stubint len=1;while(father[i]!=i){i=father[i];len++;}b[0]=i;b[1]=len;}}