Network of Schools
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 10500 | Accepted: 4189 |
Description
A number of schools are connected to a computer
network. Agreements have been developed among those schools: each school
maintains a list of schools to which it distributes software (the “receiving
schools”). Note that if B is in the distribution list of school A, then A does
not necessarily appear in the list of school B
You are to write a
program that computes the minimal number of schools that must receive a copy of
the new software in order for the software to reach all schools in the network
according to the agreement (Subtask A). As a further task, we want to ensure
that by sending the copy of new software to an arbitrary school, this software
will reach all schools in the network. To achieve this goal we may have to
extend the lists of receivers by new members. Compute the minimal number of
extensions that have to be made so that whatever school we send the new software
to, it will reach all other schools (Subtask B). One extension means introducing
one new member into the list of receivers of one school.
Input
The first line contains an integer N: the number
of schools in the network (2 <= N <= 100). The schools are identified by
the first N positive integers. Each of the next N lines describes a list of
receivers. The line i+1 contains the identifiers of the receivers of school i.
Each list ends with a 0. An empty list contains a 0 alone in the line.
Output
Your program should write two lines to the
standard output. The first line should contain one positive integer: the
solution of subtask A. The second line should contain the solution of subtask
B.
Sample Input
5
2 4 3 0
4 5 0
0
0
1 0
Sample Output
1
2
Source
强连通缩点，第一问是求入度为0的点，第二问是求入度为0和出度为0的点数 的最大值
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 #include <stack>
6
7 using namespace std;
8
9 const int MAX_N = 105;
10 int first[MAX_N],v[MAX_N * MAX_N],Next[MAX_N * MAX_N];
11 int low[MAX_N],pre[MAX_N],cmp[MAX_N];
12 int ind[MAX_N],oud[MAX_N];
13 int N;
14 int dfs_clock,scc_cnt;
15 stack<int > S;
16
17
18 void dfs(int u) {
19 low[u] = pre[u] = ++dfs_clock;
20 S.push(u);
21 for(int e = first[u]; e != -1; e = Next[e]) {
22 if(!pre[ v[e] ]) {
23 dfs(v[e]);
24 low[ u] = min(low[u],low[ v[e] ]);
25 } else {
26 if(!cmp[ v[e] ]) {
27 low[u] = min(low[u],pre[ v[e] ]);
28 }
29 }
30 }
31
32 if(low[u] == pre[u]) {
33 ++scc_cnt;
34 for(;;) {
35 int x = S.top(); S.pop();
36 cmp[x] = scc_cnt;
37 if(x == u) break;
38 }
39 }
40 }
41 void scc() {
42 dfs_clock = scc_cnt = 0;
43 memset(cmp,0,sizeof(cmp));
44 memset(pre,0,sizeof(pre));
45
46 for(int i = 1; i <= N; ++i) if(!pre[i]) dfs(i);
47 }
48
49 void add_edge(int id,int u) {
50 int e = first[u];
51 Next[id] = e;
52 first[u] = id;
53 }
54
55 void solve() {
56 scc();
57 for(int i = 1; i <= N; ++i) {
58 for(int e = first[i]; e != -1; e = Next[e]) {
59 if(cmp[ v[e] ] == cmp[i]) continue;
60 ind[ cmp[ v[e] ] ]++;
61 oud[ cmp[i] ]++;
62 }
63 }
64 int in = 0,ou = 0;
65 for(int i = 1; i <= scc_cnt; ++i) {
66 if(!ind[i]) in++;
67 if(!oud[i]) ou++;
68 }
69 printf("%d\n%d\n",in,scc_cnt == 1 ? 0 : max(in,ou));
70
71 }
72 int main()
73 {
74 //freopen("sw.in","r",stdin);
75
76 scanf("%d",&N);
77 int len = 0;
78 for(int i = 1; i <= N; ++i) first[i] = -1;
79
80 for(int i = 1; i <= N; ++i) {
81 int ch;
82 scanf("%d",&ch);
83 while(ch != 0) {
84 // printf("%d ",ch);
85
86 add_edge(len,i);
87 v[len++] = ch;
88 scanf("%d",&ch);
89 }
90 //cout << endl;
91 }
92
93 solve();
94 //cout << "Hello world!" << endl;
95 return 0;
96 }