POJ1236解题报告 强连通分支+缩点

Network of Schools

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 4177 Accepted: 1636

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：有若干个学校，每个学校可以向某些学校提供软件，第一问：最初至少应该发给几个学校软件才能保证所有学校都能得到软件？

第二问：至少要新添加几条网络才能将关系图变为强连通图？

tarjan第三题

思路：第一个问比较简单，用tarjan求出强连通分量缩点后，求入度为0的缩点数，第二问没想出来，看别人的解题报告才知道求的是

0入度缩点和0出度缩点较大的值。。太菜了~

 #include<iostream>using namespace std;struct L{       int v;   L *next;      };L *head[110];int stack[200],top=0;int dfn[110],low[110],t,suo[110],C;bool in[110];void tarjan(int v){dfn[v]=low[v]=++t;stack[top++]=v;in[v]=true;for(L *p=head[v];p!=NULL;p=p->next)if(!dfn[p->v]){tarjan(p->v);if(low[p->v]<low[v])low[v]=low[p->v];}else if(in[p->v]&&low[p->v]<low[v])low[v]=low[p->v];if(dfn[v]==low[v]){C++;do{v=stack[--top];in[v]=false;suo[v]=C;}while(low[v]!=dfn[v]);}}int main(){    int n,i,a,b;    while(cin>>n&&n){memset(head,0,sizeof(int)*110);        for(i=1;i<=n;i++)while(cin>>a&&a){L *p=new L;p->next=head[i];p->v=a;head[i]=p;}memset(dfn,0,sizeof(int)*110);memset(in,0,sizeof(bool)*110);for(i=1,C=t=0;i<=n;i++)if(!dfn[i])tarjan(i);bool f[110][110];int chudu[110],rudu[110];memset(f,true,sizeof(f));memset(chudu,0,sizeof(int)*110);memset(rudu,0,sizeof(int)*110);for(i=1;i<=n;i++)for(L *p=head[i];p!=NULL;p=p->next)if(suo[i]!=suo[p->v]&&f[suo[i]][suo[p->v]]){chudu[suo[i]]++;rudu[suo[p->v]]++;f[suo[i]][suo[p->v]]=false;}for(a=0,b=0,i=1;i<=C;i++){if(!rudu[i])a++;if(!chudu[i])b++;}cout<<a<<endl;if(C>1){i=a;if(b>i)i=b;cout<<i<<endl;}elsecout<<'0'<<endl;}}