zoukankan      html  css  js  c++  java
  • POJ The Suspects

    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

    题意

      查找与0在一个集合的人有多少个(最后的结果算上0);

    解题思路

      找根与0的根相同的;

    代码如下

     1 #include<iostream>
     2 using namespace std;
     3 const int maxn = 30100;
     4 int f[maxn], n, m;
     5 void init(){
     6     for(int i = 0; i <= n; i++)    f[i] = i;
     7 }
     8 int getf(int v){
     9     if(f[v] == v)   return v;
    10     else    return f[v] = getf(f[v]);
    11 }
    12 void meg(int v, int u){
    13     int t1 = getf(v), t2 = getf(u);
    14     if(t1 != t2){
    15         f[t2] = t1;
    16     }
    17 }
    18 int main(){
    19     while(cin >> n >> m){
    20         init();
    21         if(n == 0 && m == 0)    break;
    22         for(int i = 0; i < m; i++){
    23             int k, fir, sec;
    24             cin >> k;
    25             cin >> fir;
    26             for(int j = 01; j < k; j++){
    27                 cin >> sec;
    28                 meg(fir, sec);
    29             }
    30         }
    31         for(int i = 1; i <= n; i++)     getf(i);
    32         int sum = 0;
    33         for(int i = 0; i <= n; i++){
    34             if(f[i] == f[0]){
    35                 sum++;
    36             }
    37         }
    38         cout << sum << endl;
    39     }
    40     return 0;
    41 }
    The Suspects
    
    
    
    
    
    
  • 相关阅读:
    Centos7LDAP LDAPadmin的完整部署记录(改良版,其它文档太多坑)
    linux weblogic11g 部署
    redis离线集群安装
    Weblogic11g 10.3部署
    jdk安装部署
    tar.xz文件如何解压
    linux-Centos7安装python3并与python2共存
    ssh免密码登录配置方法
    Docker容器安装weblogic详细教程
    linux命令分块总结---多操作才是真理
  • 原文地址:https://www.cnblogs.com/zoom1109/p/11025550.html
Copyright © 2011-2022 走看看