Network of Schools
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 16343 | Accepted: 6484 |
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.
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.
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.
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
【分析】第一问就是问最少有多少个连通块，可以用Tarjan缩点，然后看入度为0的连通分量有多少个就行了。
第二问就是问最少添加几条边使得这个无向图变成一个连通块，缩点后，令father为树根数，令son为子叶数，则答案为max（father，son），若缩点后就一个点，则输出1 0。
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <time.h>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define inf 0x3f3f3f3f
#define mod 10000
typedef long long ll;
using namespace std;
const int N=105;
const int M=10005;
int s,t,n,m,cnt,tim,top,cut;
int head[N],dfn[N],low[N],stack1[N];
int num[N],in[N],out[N],vis[N];
bool flag=false;
struct man
{
int to,nxt;
}edg[M];
void addedg(int u,int v)
{
edg[cnt].to=v;edg[cnt].nxt=head[u];head[u]=cnt++;
}
void init()
{
cnt=0;tim=1;top=cut=0;
memset(head,-1,sizeof head);
memset(dfn,0,sizeof dfn);
memset(low,0,sizeof low);
memset(stack1,0,sizeof stack1);
memset(num,0,sizeof num);
memset(in,0,sizeof in);
memset(out,0,sizeof out);
memset(vis,0,sizeof vis);
memset(edg,0,sizeof edg);
}
void dfs(int u) {
int v;
low[u] = dfn[u] = ++tim;
stack1[top++] = u;
vis[u] = 1;
for(int e = head[u]; e != -1; e = edg[e].nxt)
{
v = edg[e].to;
if(!dfn[v])
{
dfs(v);
low[u] = min(low[u], low[v]);
}
else if(vis[v])
{
low[u] = min(low[u], dfn[v]);
}
}
if(low[u] == dfn[u])
{
cut++;//printf("!!!%d\n",u);
do
{
v = stack1[--top];
num[v] = cut;
vis[v] = 0;
}while(u != v);
}
}
int main() {
while(~scanf("%d",&n)){
init();
for(int i=1;i<=n;i++){
while(1){
scanf("%d",&m);
if(!m)break;
addedg(i,m);
}
}
for(int i=1;i<=n;i++)if(!dfn[i])dfs(i);
for(int i=1;i<=n;i++){
for(int j=head[i];j!=-1;j=edg[j].nxt){
int v=edg[j].to;//printf("%d %d %d %d\n",i,num[i],v,num[v]);
if(num[i]!=num[v])out[num[i]]++,in[num[v]]++;
}
}
int father=0,son=0;
for(int i=1;i<=cut;i++){
if(in[i]==0)father++;
if(out[i]==0)son++;
}
if(cut==1)printf("1\n0\n");
else printf("%d\n%d\n",father,max(father,son));
}
return 0;
}