POJ 1236 Network of Schools

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最新推荐文章于 2024-02-06 19:17:23 发布

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原文链接：http://www.cnblogs.com/crackpotisback/p/3935359.html

本文介绍了一种基于强连通缩点算法解决软件分发问题的方法，通过计算入度为0和出度为0的块数来确定最少的软件分发学校数量及网络扩展需求。

Network of Schools

Time Limit: 1000ms

Memory Limit: 10000KB

This problem will be judged on PKU. Original ID: 1236
64-bit integer IO format: %lld Java class name: Main

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

Sample Output

1
2

Source

IOI 1996

解题：强连通缩点，求入度为0的块数，和出度为0的块数。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <climits>
 7 #include <vector>
 8 #include <queue>
 9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 using namespace std;
17 const int maxn = 110;
18 vector<int>g[maxn];
19 stack<int>stk;
20 int dfn[maxn],low[maxn],belong[maxn],cnt,scc;
21 int n,in[maxn],out[maxn];
22 bool instack[maxn];
23 void tarjan(int u){
24     dfn[u] = low[u] = ++cnt;
25     instack[u] = true;
26     stk.push(u);
27     for(int i = 0; i < g[u].size(); i++){
28         if(!dfn[g[u][i]]){
29             tarjan(g[u][i]);
30             low[u] = min(low[u],low[g[u][i]]);
31         }else if(dfn[g[u][i]] && instack[g[u][i]]) low[u] = min(low[u],dfn[g[u][i]]);
32     }
33     if(dfn[u] == low[u]){
34         scc++;
35         int v;
36         do{
37             v = stk.top();
38             instack[v] = false;
39             stk.pop();
40             belong[v] = scc;
41         }while(v != u);
42     }
43 }
44 int main(){
45     int i,j,v;
46     while(~scanf("%d",&n)){
47         for(i = 0; i <= n; i++){
48             dfn[i] = low[i] = belong[i] = 0;
49             in[i] = out[i] = 0;
50             instack[i] = false;
51             g[i].clear();
52         }
53         while(!stk.empty()) stk.pop();
54         cnt = scc = 0;
55         for(i = 1; i <= n; i++)
56             while(scanf("%d",&v),v) g[i].push_back(v);
57         for(i = 1; i <= n; i++) if(!dfn[i]) tarjan(i);
58         for(i = 1; i <= n; i++){
59             for(j = 0; j < g[i].size(); j++){
60                 if(belong[i] != belong[g[i][j]]){
61                     in[belong[g[i][j]]]++;
62                     out[belong[i]]++;
63                 }
64             }
65         }
66         int ans1 = 0,ans2 = 0;
67         for(i = 1; i <= scc; i++){
68             if(!in[i]) ans1++;
69             if(!out[i]) ans2++;
70         }
71         printf("%d\n%d\n",ans1,scc == 1?0:max(ans1,ans2));
72     }
73     return 0;
74 }