Poj 1611 的傳送門

***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

題目大意：第一行為兩個整數n和m, 其中n是學生的數量, m是團體的數量。0 < n <= 30000，0 <= m <= 500。
每個學生編號是一個0到n-1之間的整數，一開始只有0號學生被視為可能的患者。
緊隨其后的是團體的成員列表，每組一行。
每一行有一個整數k，代表成員數量。之后,有k個整數代表這個群體的學生。一行中的所有整數由至少一個空格隔開。
n = m = 0表示輸入結束，不需要處理。
然后輸出可能感染的人的個數

解題思路：并查集：

具體詳見代碼：

#include <iostream> #include <cstdio> using namespace std; const int maxn = 30010;int fa[maxn];//父親節點 int rank[maxn];//rank[x]就是表示x的高度上的一個上界 int sum[maxn];//sum[]存儲該集合中元素個數，并在集合合并時更新sum[]即可 void Init(int x)//創建單元集 {fa[x]=x;sum[x]=1;rank[x]=0;return ; }int Find(int x)//遞歸找父親，路徑壓縮 {if(x != fa[x])fa[x]=Find(fa[x]);return fa[x]; }void Union(int a, int b)//讓rank比較高的作為父親節點 {a=Find(a);b=Find(b);if(a == b)return ;if(rank[a] > rank[b]){fa[b]=a;sum[a]+=sum[b];}else{fa[a]=b;if(rank[a] == rank[b])rank[b]++;sum[b]+=sum[a];} } int main() {int m,n,k,a,b;while(cin>>m>>n){if(!m && !n)break;if(n == 0){puts("1");continue;}for(int i=0; i<m; i++)Init(i);while(n--){cin>>k>>a;k--;while(k--){cin>>b;Union(a, b);}}cout<<sum[Find(0)]<<endl;}return 0; } 創作挑戰賽新人創作獎勵來咯，堅持創作打卡瓜分現金大獎

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