B - The Suspects POJ - 1611

Posted 2021-03-12 orangeko

B - The Suspects

POJ - 1611

Time Limit: 1000MS		Memory Limit: 20000K
Total Submissions: 61692		Accepted: 29146

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

Source

Asia Kaohsiung 2003

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思路：

　　维护一个小根堆的并查集，最后O(n)过一遍就可以了

 

CODE

 

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <algorithm>
  4 #include <iostream>
  5 #include <cmath>
  6 #include <assert.h>
  7 #include <vector>
  8 
  9 #define dbg(x) cout << #x << "=" << x << endl
 10 
 11 using namespace std;
 12 typedef long long LL;
 13 const int maxn = 30007;
 14 
 15 int fa[maxn];
 16 int n,m,ans,cnt;
 17 bool vis[maxn];
 18 int a[maxn];
 19 
 20 //vector <int> v;
 21 
 22 void init()
 23 {
 24     for(int i = 1; i <= n; i++) {
 25         fa[i] = i;
 26     }
 27 }
 28 
 29 int fid(int x)
 30 {
 31     int r = x;
 32     while(fa[r] != r) {
 33         r = fa[r];
 34     }
 35     int i,j;///路径压缩
 36     i = x;
 37     while(fa[i] != r) {
 38         j = fa[i];
 39         fa[i] =  r;
 40         i = j;
 41     }
 42     return r;
 43 }
 44 
 45 void join(int r1, int r2)///合并
 46 {
 47     int fidroot1 = fid(r1), fidroot2 = fid(r2);
 48     int root = min(fidroot1, fidroot2);
 49     if(fidroot1 != fidroot2) {
 50         fa[fidroot1] = root;
 51         fa[fidroot2] = root;
 52     }
 53 }
 54 
 55 template<class T>inline void read(T &res)
 56 {
 57     char c;T flag=1;
 58     while((c=getchar())<‘0‘||c>‘9‘)if(c==‘-‘)flag=-1;res=c-‘0‘;
 59     while((c=getchar())>=‘0‘&&c<=‘9‘)res=res*10+c-‘0‘;res*=flag;
 60 }
 61 
 62 namespace _buff {
 63     const size_t BUFF = 1 << 19;
 64     char ibuf[BUFF], *ib = ibuf, *ie = ibuf;
 65     char getc() {
 66         if (ib == ie) {
 67             ib = ibuf;
 68             ie = ibuf + fread(ibuf, 1, BUFF, stdin);
 69         }
 70         return ib == ie ? -1 : *ib++;
 71     }
 72 }
 73 
 74 int qread() {
 75     using namespace _buff;
 76     int ret = 0;
 77     bool pos = true;
 78     char c = getc();
 79     for (; (c < ‘0‘ || c > ‘9‘) && c != ‘-‘; c = getc()) {
 80         assert(~c);
 81     }
 82     if (c == ‘-‘) {
 83         pos = false;
 84         c = getc();
 85     }
 86     for (; c >= ‘0‘ && c <= ‘9‘; c = getc()) {
 87         ret = (ret << 3) + (ret << 1) + (c ^ 48);
 88     }
 89     return pos ? ret : -ret;
 90 }
 91 
 92 int main()
 93 {
 94     while(scanf("%d %d",&n, &m) && (n+m)) {
 95         int ans = 0;
 96         if(m == 0) {
 97             puts("1");
 98             continue;
 99         }
100         init();
101         //v.clear();
102         while(m--) {
103             int k;
104             scanf("%d",&k);
105             memset(a,0,sizeof(a));
106             for(int i = 1; i <= k; ++i) {
107                 scanf("%d",&a[i]);
108                 if(i == 1) continue;
109                 join(a[i], a[i-1]);
110             }
111         }
112         int pre = fid(0);
113         for(int i = 0; i < n; ++i) {
114             int now = fid(i);
115             //printf("fa[%d]:%d
",i,now);
116             if(now == pre) {
117                 ans++;
118             }
119         }
120         cout << ans << endl;
121     }
122     return 0;
123 }

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