poj 1611:The Suspects（並查集，經典題）

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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 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 　　 並查集，經典題。　　 並查集模板（來自北大課件）：

 1 int ufs[MAXN];    //並查集 2  3 void Init(int n)    //初始化 4 { 5     int i; 6     for(i=0;i<n;i++){ 7         ufs[i] = i; 8     } 9 }10 11 int GetRoot(int a)    //獲得a的根節點。路徑壓縮12 {13     if(ufs[a]!=a){    //沒找到根節點14         ufs[a] = GetRoot(ufs[a]);15     }16     return ufs[a];17 }18 19 void Merge(int a,int b)    //合并a和b的集合20 {21     ufs[GetRoot(b)] = GetRoot(a);22 }23 24 bool Query(int a,int b)    //查詢a和b是否在同一集合25 {26     return GetRoot(a)==GetRoot(b);27 }

 

　　 題意：　　 　　 思路：　　很經典的並查集的題目，找一個sum[]數組記錄每一個以當前下標為根節點的集合的個體數目，最後輸出0號的根節點對應的sum值，就是0號學生所在團體的人數。 　　 代碼：

 1 #include <iostream> 2 #include <stdio.h> 3 using namespace std; 4 #define MAXN 30010  5 int sum[MAXN];    //集合總數 6 int ufs[MAXN];    //並查集 7  8 void Init(int n)    //初始化 9 {10     int i;11     for(i=0;i<n;i++){12         ufs[i] = i;13         sum[i] = 1;14     }15 }16 17 int GetRoot(int a)    //獲得a的根節點。路徑壓縮18 {19     if(ufs[a]!=a){    //沒找到根節點20         ufs[a] = GetRoot(ufs[a]);21     }22     return ufs[a];23 }24 25 void Merge(int a,int b)    //合并a和b的集合26 {27     int x = GetRoot(a);28     int y = GetRoot(b);29     if(x!=y){30         ufs[y] = x;31         sum[x] += sum[y];32     }33 }34 35 int main()36 {37     int n,m;38     while(scanf("%d%d",&n,&m)!=EOF){39         if(n==0 && m==0) break;40         Init(n);    //初始化並查集41         while(m--){    //讀入m行42             int t,one,two;43             scanf("%d",&t);    //每一行有t個數需要輸入44             scanf("%d",&one);45             t--;46             while(t--){47                 scanf("%d",&two);48                 Merge(one,two);    //合并集合49             }50         }51         printf("%d\n",sum[GetRoot(0)]);52     }53     return 0;54 }

 

Freecode : www.cnblogs.com/yym2013