题目链接：
http://poj.org/problem?id=1236
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Network of Schools
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 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 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 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 Source |
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题目意思：
给出每个学校能够传送信息的学校名单。
求Subtask A:至少要给多少个学校发送源信息，才能是所有学校都收到信息。求Subtask B：求至少需要添加几个接受信息学校，使得当信息发送给任意一个学校时，都能传送到其他学校。
解题思路：
先求有向图的强连通分量。然后缩点。统计各点出度和入度。
Subtask A 就是求入度为0的点的个数。
Subtask B就是求max(入度为0个数，出度为零个数）。
把每个出度为零的节点连到下一颗树的入度为0的节点上。
代码：
//#include<CSpreadSheet.h>
#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#include<cmath>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
#define Maxn 110
int low[Maxn],dfn[Maxn],dindex,n;
int sta[Maxn],belong[Maxn],bcnt,ss;
bool iss[Maxn];
int de1[Maxn],de2[Maxn];
vector<vector<int> >myv;
void tarjan(int cur)
{
//printf(":%d\n",cur);
//system("pause");
int ne;
dfn[cur]=low[cur]=++dindex;
iss[cur]=true;
sta[++ss]=cur;
for(int i=0;i<myv[cur].size();i++)
{
ne=myv[cur][i];
if(!dfn[ne])
{
tarjan(ne);
low[cur]=min(low[cur],low[ne]);
}
else if(iss[ne]&&dfn[ne]<low[cur])
low[cur]=dfn[ne];
}
if(dfn[cur]==low[cur])
{
bcnt++;
do
{
ne=sta[ss--];
iss[ne]=false;
belong[ne]=bcnt;
}while(ne!=cur);
}
}
void solve()
{
int i;
ss=bcnt=dindex=0;
memset(dfn,0,sizeof(dfn));
memset(iss,false,sizeof(iss));
for(int i=1;i<=n;i++)
if(!dfn[i])
tarjan(i);
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
while(~scanf("%d",&n))
{
myv.clear();
myv.resize(n+1);
for(int i=1;i<=n;i++)
{
int a;
while(scanf("%d",&a)&&a)
myv[i].push_back(a);
}
solve();
if(bcnt==1) //本来就是一个强连通分量
{
printf("1\n0\n");
continue;
}
int ansa=0,ansb=0;
memset(de1,0,sizeof(de1));
memset(de2,0,sizeof(de2));
for(int i=1;i<=n;i++)
{
for(int j=0;j<myv[i].size();j++)
{
int ne=myv[i][j];
if(belong[i]!=belong[ne])
{
de1[belong[ne]]++;//入度
de2[belong[i]]++; //出度
}
}
}
for(int i=1;i<=bcnt;i++)
{
if(!de1[i])
ansa++;
if(!de2[i])
ansb++;
}
ansb=max(ansb,ansa);
printf("%d\n%d\n",ansa,ansb);
}
return 0;
}