POJ1611-The Suspects-白红宇

POJ1611-The Suspects

阅读量：6606 次

发布时间：2019-06-24

本文共 2948 字，大约阅读时间需要 9 分钟。

题目链接：

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

思路：我刚开始想的复杂，还有就是输入数据也会出错。这里用了sum数组。把每一个点当作这个集合的人数，初始化为自己本身。合并的时候，把人也合并。

AC代码：

#include
    
     #include
     
      #include
      
       #include
       
        #include
        
         #include
         #include
          
           #include
           
            #include
            
             #include
             
              #include
              
               using namespace std;const int MAX = 30010;const int INF = 0X3f3f3f;int father[MAX];int sum[MAX];int n, m;int vis[MAX];void init() { for(int i = 0; i < n; i++) {//卡了我半天， 初始化用n 我原来用的MAX。 father[i] = i; sum[i] = 1; }}int findfather(int x) {//路径压缩 int a = x; while(x != father[x]) { x = father[x]; } while(a != father[a]) { int z = a; a = father[a]; father[z] = a; } return x;}void Union(int a, int b) {//合并 int faA = findfather(a); int faB = findfather(b); if(faA != faB) { father[faA] = faB; sum[faB] += sum[faA];//注意不要合并错了 }}int main() { while(scanf("%d %d", &n, &m) != EOF) { if(n == 0 && m == 0) break; init(); while(m--) {//并查集，常用的处理数据方式 int k, u, v; scanf("%d", &k); scanf("%d", &u); k--; while(k--) { scanf("%d", &v); Union(u, v); } } printf("%d\n", sum[findfather(0)]);//找到0的祖先，看这个集合多少人 } return 0;}

转载于:https://www.cnblogs.com/ACMerszl/p/9572996.html

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