B - The Suspects
| 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
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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\n",i,now);
116 if(now == pre) {
117 ans++;
118 }
119 }
120 cout << ans << endl;
121 }
122 return 0;
123 }
原文地址：https://www.cnblogs.com/orangeko/p/12264121.html