POJ 1236:Network of Schools

Network of Schools

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 12959 Accepted: 5176

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

52 4 3 04 5 0001 0

Sample Output

12

题意是给你N个学校，然后给你每一个学校能往哪些其他学校传输文件的关系。

求两个问题：1.计算如果要每个学校都能有文件的话，文件至少要准备几份。

2.如果一份文件传给任意一个学校都能保证所有学校都得到文件的话，那么该学校关系图还需要添加多少条关系。

Tarjan缩点之后

1.求入度为0的点的个数。

2.求入度为0和出度为0的点的个数的最大值。

代码：

#include <iostream>  #include <string>#include <cstring>#include <queue>  #pragma warning(disable:4996)using namespace std;  int head[10005],LOW[10005],DFN[10005],instack[10005],Stack[10005],Belong[10005],out[10005],in[10005];int n,m,edge_num,Dindex,Stop,Bcnt;struct edge{int to;int next;}Edge[50005];void init(){edge_num=0;Stop=Bcnt=Dindex=0;memset(Edge,-1,sizeof(Edge));memset(head,-1,sizeof(head));memset(LOW,0,sizeof(LOW));memset(DFN,0,sizeof(DFN));memset(instack,0,sizeof(instack));memset(Stack,0,sizeof(Stack));memset(Belong,0,sizeof(Belong));memset(out,0,sizeof(out));memset(in,0,sizeof(in));}void addedge(int u,int v){Edge[edge_num].to=v;Edge[edge_num].next=head[u];head[u]=edge_num;edge_num++;}void tarjan(int i){int j;DFN[i]=LOW[i]=++Dindex;instack[i]=true;Stack[++Stop]=i;for(j=head[i];j!=-1;j=Edge[j].next){int v=Edge[j].to;if(DFN[v]==0){tarjan(v);LOW[i]=min(LOW[i],LOW[v]);}else if(instack[v]==1){LOW[i]=min(LOW[i],DFN[v]);}}if(DFN[i]==LOW[i]){Bcnt++;do{j=Stack[Stop--];instack[j]=false;Belong[j]=Bcnt;}while(j!=i);}}void solve(){int i,j,u,v;cin>>n;init();m=0;for(i=1;i<=n;i++){while(1){cin>>v;if(v==0)break;addedge(i,v);m++;}}for(i=1;i<=n;i++){if(!DFN[i])tarjan(i);}for(i=1;i<=n;i++){for(j=head[i];j!=-1;j=Edge[j].next){if(Belong[i]!=Belong[Edge[j].to])in[Belong[Edge[j].to]]++;//计算缩点后每个点的入度}}for(i=1;i<=n;i++){for(j=head[i];j!=-1;j=Edge[j].next){if(Belong[i]!=Belong[Edge[j].to])out[Belong[i]]++;//计算缩点后每个点的出度}}int result1=0,result2=0;for(i=1;i<=Bcnt;i++){if(out[i]==0){result1++;}}for(i=1;i<=Bcnt;i++){if(in[i]==0){result2++;}}if(Bcnt==1)/////////////////////注意若是连通块只有1个的时候，输出 1 0，不需要添加边了！！！{cout<<1<<endl;cout<<0<<endl;}else{cout<<result2<<endl;cout<<max(result1,result2)<<endl;}}int main()  {  //freopen("i.txt","r",stdin);//freopen("o.txt","w",stdout);solve();return 0;  }