Network of Schools
IOI '96 Day 1 Problem 3
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.
PROGRAM NAME: schlnet
INPUT FORMAT
The first line of the input file 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.
SAMPLE INPUT (file schlnet.in)
5
2 4 3 0
4 5 0
0
0
1 0
OUTPUT FORMAT
Your program should write two lines to the output file. The first line should contain one positive integer: the solution of subtask A. The second line should contain the solution of subtask B.
SAMPLE OUTPUT (file schlnet.out)
1
2
--------------------------------------------------------------------------------
算法：強聯通分支，雙聯通分支
首先使用強聯通分子算法，把同一個分支變成一個點，然後把有向圖當成無向圖，求得雙聯通分支。
求出每一個聯通分量的入度爲0的點和出度爲0的點。
task A很好求，只需要把所有的入度爲0的點加起來就是了；如果某個聯通分量沒有入度爲0 的點，那麼就加1.
task B要求把這些聯通分量連接成一個連通圖，最簡單的辦法是把這些聯通分量連成環。
假第i個聯通分量出度爲0的點有out個，第i+1個聯通分量入度爲0的點有in個，那麼最少需要max（out，in）個遍才能把他們連接起來。
由於要求最少的遍，那麼當out和in最接近的時候，能節省很多邊，
考慮這個例子（in,out):(1,2)，（1,1),(2,1);如果這樣按順序連接起來的話，需要2+1+2+1條邊，但如果2和3顛倒一下順序，則需要1+2+1+1條邊
因此，每次選取還未匹配的分量 出度最大的一個分量和入度最大的另一個分量，連接起來。
這題好麻煩，很多需要注意的問題。
這題是用java寫的，運行速度比c++慢多了
Executing...
Test 1: TEST OK [0.090 secs, 254972 KB]
Test 2: TEST OK [0.090 secs, 254972 KB]
Test 3: TEST OK [0.108 secs, 254972 KB]
Test 4: TEST OK [0.126 secs, 254972 KB]
Test 5: TEST OK [0.126 secs, 254972 KB]
Test 6: TEST OK [0.126 secs, 255384 KB]
Test 7: TEST OK [0.144 secs, 255248 KB]
Test 8: TEST OK [0.162 secs, 255420 KB]
Test 9: TEST OK [0.270 secs, 255792 KB]
Test 10: TEST OK [0.144 secs, 255556 KB]
Test 11: TEST OK [0.162 secs, 255240 KB]
]-----------------------------------------------------------------------------------
/*
ID:yunleis3
PROG:schlnet
LANG:JAVA
*/
import java.io.File;
import java.io.FileWriter;
import java.io.IOException;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.Scanner;
public class schlnet {
private static final ArrayList<Integer> outdgree0 = null;
static ArrayList<Integer>[] list;
static ArrayList<Integer>[] reverse_list;
static int current_cpnt=0;
static int []dfs_nmbr;
static int []cpnt_nmbr;
static int []dfs_high;
static int dfs_n;
static LinkedList<Integer> dfs_stack=new LinkedList<Integer>();
public static void main(String [] arg) throws IOException{
File file=new File("schlnet.in");
long begin=System.currentTimeMillis();
Scanner scan=new Scanner(file);
FileWriter w=new FileWriter(new File("schlnet.out"));
int total =scan.nextInt();
list=new ArrayList [total+1];
reverse_list=new ArrayList [total+1];
int out=0;
int in=0;
int[] indegree=new int [total+1];
for(int i=1;i<total+1;i++){reverse_list[i]=new ArrayList<Integer>();}
for(int i=1;i<total+1;i++){
list[i]=new ArrayList<Integer>();
while(true){
int t=scan.nextInt();
if(t==0) break;
list[i].add(t);
//indegree[t]++;
reverse_list[t].add(i);
}
if(list[i].size()==0) out++;
}
dfs_nmbr=new int[total+1];
cpnt_nmbr=new int[total+1];
dfs_high=new int[total+1];
dfs_n=total+1;
for(int i=1;i<=total;++i){
if(dfs_nmbr[i]==0)
scc(i);
}
ArrayList<Integer>[] list1=new ArrayList[current_cpnt+1];
ArrayList<Integer>[] reverse_list1=new ArrayList[current_cpnt+1];
for(int i=1;i<current_cpnt+1;++i){
list1[i]=new ArrayList<Integer>();
reverse_list1[i]=new ArrayList<Integer>();
}
for(int i=1;i<=total;i++){
int first=cpnt_nmbr[i];
for(int j=0;j<list[i].size();++j){
int next=cpnt_nmbr[list[i].get(j)];
if(next==first)
continue;
boolean flag=false;
for(int s=0;s<list1[first].size();++s){
if(list1[first].get(s)==next)
{
flag=true;
break;
}
}
if(!flag)
{
list1[first].add(next);
reverse_list1[next].add(first);
indegree[next]++;
}
}
}
list=list1;
reverse_list=reverse_list1;
total=current_cpnt;
for(int i=1;i<=total;i++){
if(indegree[i]==0) in++;
}
boolean[] visited=new boolean[total+1];
int [] componentnum=new int[total+1];
for(int i=0;i<total+1;++i)
{
visited[i]=false;
componentnum[i]=-1;
componentnum[i]=-1;
}
LinkedList<Integer> stack=new LinkedList<Integer>();
int cpnt_g=0;
ArrayList<Integer> indgrees0=new ArrayList<Integer>();
ArrayList<Integer> outdgrees0=new ArrayList<Integer>();
while(true){
boolean find0indgree=false;
int root=0;
boolean finished=true;
ArrayList<Integer> nums=new ArrayList<Integer>();
for(int i=1;i<total+1;++i){
if(!visited[i]&&indegree[i]==0){
root=i;
find0indgree=true;
}
if(!visited[i])
finished=false;
}
if(finished)
break;
if(!find0indgree){
for(int i=1;i<total+1;++i)
if(!visited[i])
root=i;
}
stack.addFirst(root);
visited[root]=true;
int cmpnt_l=cpnt_g++;
nums.add(root);
componentnum[root]=cmpnt_l;
while(!stack.isEmpty()){
int r=stack.removeFirst();
for(int i=0,size=list[r].size();i<size;++i){
int next=list[r].get(i);
if(!visited[next])
{
visited[next]=true;
stack.addFirst(next);
componentnum[next]=cmpnt_l;
nums.add(next);
}
}
for(int i=0,size=reverse_list[r].size();i<size;++i){
int next=reverse_list[r].get(i);
if(!visited[next])
{
visited[next]=true;
stack.addFirst(next);
componentnum[next]=cmpnt_l;
nums.add(next);
}
}
}
int ins0=0,outs0=0;
for(int i=0;i<nums.size();++i){
if(indegree[nums.get(i)]==0)
++ins0;
if(list[nums.get(i)].size()==0)
++outs0;
}
indgrees0.add(ins0);
outdgrees0.add(outs0);
}
int taska=0,taskb=0;
if(total==1){
taska=1;
taskb=0;
}
else if(cpnt_g==1&&indgrees0.get(0)==0&&outdgrees0.get(0)==0){
taska=1;
taskb=0;
}else if(cpnt_g==1){
taska=indgrees0.get(0);
if(taska==0)
taska=1;
taskb=indgrees0.get(0)>outdgrees0.get(0)?indgrees0.get(0):outdgrees0.get(0);
}
else{
in=0;out=0;
boolean[] outvisited=new boolean[cpnt_g];
boolean[] invisited=new boolean[cpnt_g];
for(int i=0;i<cpnt_g;++i){
//pick the max out;
int ptr=-1;
for(int j=0;j<cpnt_g;++j){
if(!outvisited[j]&&(ptr==-1||outdgrees0.get(ptr)>outdgrees0.get(ptr)))
ptr=j;
}
int ptrin=-1;
for(int j=0;j<cpnt_g;++j){
if(!invisited[j]&&(ptrin==-1||indgrees0.get(ptrin)>indgrees0.get(ptrin)))
ptrin=j;
}
outvisited[ptr]=true;
invisited[ptrin]=true;
in=indgrees0.get(ptrin);
out=outdgrees0.get(ptr);
if(in==0)
in=1;
if(out==0)
out=1;
taska+=in;
taskb+=max(out,in);
}
// for(int i=0;i<cpnt_g;++i){
// int next=(i+1)%cpnt_g;
// out=outdgrees0.get(i);
// if(out==0)
// out=1;
// in=indgrees0.get(next);
// if(in==0)
// in=1;
// taska+=in;
// taskb+=max(out,in);
// }
}
FileWriter wr=new FileWriter(new File("schlnet.out"));
wr.write(taska+"\n"+taskb+"\n");
wr.flush();
wr.close();
System.out.print(taska+"\n"+taskb+"\n");
}
static void scc(int v){
dfs_nmbr[v]=dfs_n--;
dfs_stack.addFirst(v);
dfs_high[v]=dfs_nmbr[v];
for(int i=0;i<list[v].size();++i){
int w=list[v].get(i);
if(dfs_nmbr[w]==0){
scc(w);
dfs_high[v]=dfs_high[v]>dfs_high[w]?dfs_high[v]:dfs_high[w];
}
else
if(dfs_nmbr[w]>dfs_nmbr[v]&&cpnt_nmbr[w]==0)
dfs_high[v]=max(dfs_high[v],dfs_nmbr[w]);
}
if(dfs_high[v]==dfs_nmbr[v]){
++current_cpnt;
while(!dfs_stack.isEmpty()){
int x=dfs_stack.removeFirst();
cpnt_nmbr[x]=current_cpnt;
if(x==v){
break;
}
}
}
}
static int max(int x,int y){
return x>y?x:y;
}
}