題目鏈接：
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
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 Source |
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題目意思：給出每個學校能夠傳送信息的學校名單。
求Subtask A:至少要給多少個學校發送源信息，才能是所有學校都收到信息。求Subtask B：求至少需要添加幾個接受信息學校，使得當信息發送給任意一個學校時，都能傳送到其他學校。
解題思路：
先求有向圖的強連通分量。然後縮點。統計各點出度和入度。
Subtask A 就是求入度爲0的點的個數。
Subtask B就是求max(入度爲0個數，出度爲零個數）。
把每個出度爲零的節點連到下一顆樹的入度爲0的節點上。
代碼：
//#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) //本來就是一個強連通分量
{
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;
}
