[强连通分量] POJ 1236 Network of Schools

Network of Schools

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 16803		Accepted: 6641

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
原题大意：给学校的数目n，接下来n行，第i行有若干个数，以0结束，代表了第i-1个学校可以传递消息的学校编号（单向），问，最少需要多少份拷贝资料，如果所有学校两两间可以传达到资料，那么需要新建多少个列表？

解题思路： 同样缩点，很容易想出入度为0的点即为最少拷贝数。
           入度为0的点和出度为0的点的最大值便是所需新建数。

#include<stdio.h>
#include<string.h>
struct list
  {
  	 int v;
  	 list *next;
  };
list *head[10010],*rear[10010];
int dfn[10010],low[10010],stack[10010],s[10010],times,cnt,top;
bool instack[10010];
void tarjian(int v)
   {
   	  dfn[v]=low[v]=++times;
   	  stack[top++]=v;
   	  instack[v]=true;
   	  for(list *p=head[v];p!=NULL;p=p->next)
   	     if(!dfn[p->v])
   	       {
   	       	  tarjian(p->v);
   	       	  if(low[p->v]<low[v]) low[v]=low[p->v];
		   }else if(instack[p->v]&&low[p->v]<low[v]) low[v]=low[p->v];
	  if(dfn[v]==low[v])
	     {
	     	++cnt;
	     	do
	     	{
	     	   v=stack[--top];
			   instack[v]=false;
			   s[v]=cnt;
			}while(dfn[v]!=low[v]);
		 }
   	  return;
   }
int main()
  {
  	 int n,k,j,i,rud[10010],chd[10010],ans1,ans2;
  	 scanf("%d",&n);
  	 memset(head,0,sizeof(head));
  	 memset(rear,0,sizeof(rear));
  	 for(j=1;j<=n;++j)
  	   {
  	   	 head[j]=rear[j]=new list;
  	   	 rear[j]->v=j;
  	   	 while(scanf("%d",&k),k)
  	   	  {
  	   	     rear[j]->next=new list;
  	   	     rear[j]=rear[j]->next;
  	   	     rear[j]->v=k;
		  }
		 rear[j]->next=NULL;
	   }
	 memset(dfn,0,sizeof(dfn));
	 memset(low,0,sizeof(low));
	 memset(s,0,sizeof(s));
	 memset(instack,false,sizeof(instack));
	 times=cnt=top=0;
	 for(i=1;i<=n;++i) if(!dfn[i]) tarjian(i);
	 if(cnt==1)
	   {
	   	  printf("1\n0\n");
	   	  return 0;
	   }
	 memset(rud,0,sizeof(rud));memset(chd,0,sizeof(chd));
	 ans1=ans2=0;
	 for(i=1;i<=n;++i)
	   for(list *p=head[i];p!=NULL;p=p->next)
	      if(s[p->v]!=s[i])
	        {
	           ++rud[s[p->v]];
	           ++chd[s[i]];
			}
	 for(i=1;i<=cnt;++i) if(!rud[i]) ++ans1;
	 for(i=1;i<=cnt;++i) if(!chd[i]) ++ans2;
	 printf("%d\n",ans1);
	 printf("%d\n",ans1>ans2?ans1:ans2);
  	 return 0;
  }