題目

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

分析與解答

通過改了一個連接函數(shù)join，我們的根的num直接儲存他所在的樹的結(jié)點個數(shù)，由于題目說0號事已經(jīng)有了，那我們找到0號的根然后輸出跟的num即可

#include <cstdio> #include <cstring> #include <algorithm> #define maxn 33000 using namespace std;int per[maxn], num[maxn], n, m; int a[maxn];int find(int x){if(x == per[x])return x;return per[x] = find(per[x]); }void join(int x, int y){int fx = find(x);int fy = find(y);if(fx != fy){per[fx] = fy;//把fy規(guī)定為fx祖宗 num[fy] = num[fx]+num[fy];//fy所在的數(shù)的結(jié)點的個數(shù)（包含他自己） }return ; }int main (){while(scanf("%d%d", &n, &m)){if(n==0&&m==0) return 0;for(int i = 0; i < n ; ++i){per[i] = i;num[i] = 1;}while(m--){int t;scanf("%d", &t);for(int i = 0; i < t; ++i){scanf("%d", &a[i]);}for(int i = 0 ; i < t - 1; ++i)join(a[i], a[i + 1]);}int k = find(0);printf("%d\n", num[k]);}return 0; }

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