POJ1236 Network of Schools

Network of Schools

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Network of Schools

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 25741		Accepted: 10194

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

IOI 1996

N(2<N<100)各学校之间有单向的网络，每个学校得到一套软件后，可以通过单向网络向周边的学校传输，问题1：初始至少需要向多少个学校发放软件，使得网络内所有的学校最终都能得到软件。2，至少需要添加几条传输线路(边)，使任意向一个学校发放软件后，经过若干次传送，网络内所有的学校最终都能得到软件。

题解

参照嘉庆帝的题解。

我们可以先tarjan求出scc，进行缩点求出dag图，然后我们考虑第一个问题，求最少发几套软件可以全覆盖，首先题意已经保证了是联通的。然后我们可以想，如果我们把所有没有入边的点都放上软件，是一定可行的。有入边的一定会通过一些边最终从一定有出边的发放软件的地方获得软件。

然后我们考虑第二个问题。这是一个连通图。如果我们有些点没有入点，有些点没出点。那我们如果想办法将入点和一些出点相连，就能保证最后会成为很多圆相连。这样子答案就是没有入边的点和没有出边的点的最大值。意思就是说把这些点连成一个环。

时间复杂度(O(n+m))

#include<iostream>
#include<vector>
#define rg register
#define il inline
#define co const
template<class T>il T read(){
    rg T data=0,w=1;rg char ch=getchar();
    for(;!isdigit(ch);ch=getchar())if(ch=='-') w=-w;
    for(;isdigit(ch);ch=getchar()) data=data*10+ch-'0';
    return data*w;
}
template<class T>il T read(rg T&x) {return x=read<T>();}
typedef long long ll;
using namespace std;

co int N=101;
int n,dfn[N],low[N],num=0;
int st[N],top=0,tot=0,c[N],ru[N],chu[N];
bool v[N];
vector<int> e[N],scc[N],sc[N];

void tarjan(int x){
	dfn[x]=low[x]=++num;
	st[++top]=x;
	v[x]=1;
	for(unsigned i=0;i<e[x].size();++i){
		int y=e[x][i];
		if(!dfn[y]){
			tarjan(y);
			low[x]=min(low[x],low[y]);
		}
		else if(v[y]) low[x]=min(low[x],dfn[y]);
	}
	if(dfn[x]==low[x]){
		v[x]=0;
		scc[++tot].push_back(x);
		c[x]=tot;
		int y;
		while(x!=(y=st[top--])){
			scc[tot].push_back(y);
			v[y]=0;
			c[y]=tot;
		}
	}
}
int main(){
	read(n);
	for(int i=1;i<=n;++i){
		for(int x;read(x);) e[i].push_back(x);
	}
	for(int i=1;i<=n;++i)
		if(!dfn[i]) tarjan(i);
	for(int x=1;x<=n;++x)
		for(unsigned i=0;i<e[x].size();++i){
			int y=e[x][i];
			if(c[x]==c[y]) continue;
			sc[c[x]].push_back(c[y]);
			++ru[c[y]];
			++chu[c[x]];
		}
	int ansa=0,ansb=0;
	for(int i=1;i<=tot;++i){
		if(!ru[i]) ++ansa;
		if(!chu[i]) ++ansb;
	}
	printf("%d
",ansa);
	if(tot==1) puts("0");
	else printf("%d
",max(ansa,ansb));
	return 0;
}