poj 1236
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Description
A numberof schools are connected to a computer network. Agreements have been developedamong those schools: each school maintains a list of schools to which itdistributes software (the “receiving schools”). Note that if B is in thedistribution list of school A, then A does not necessarily appear in the listof school B You are to write a program that computes the minimal number of schools thatmust receive a copy of the new software in order for the software to reach allschools in the network according to the agreement (Subtask A). As a furthertask, we want to ensure that by sending the copy of new software to anarbitrary school, this software will reach all schools in the network. Toachieve 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 whateverschool we send the new software to, it will reach all other schools (SubtaskB). One extension means introducing one new member into the list of receiversof one school.
Input
The firstline contains an integer N: the number of schools in the network (2 <= N<= 100). The schools are identified by the first N positive integers. Eachof the next N lines describes a list of receivers. The line i+1 contains theidentifiers of the receivers of school i. Each list ends with a 0. An emptylist contains a 0 alone in the line.
Output
Yourprogram should write two lines to the standard output. The first line shouldcontain one positive integer: the solution of subtask A. The second line shouldcontain the solution of subtask B.
Sample Input
5
2 4 3 0
4 5 0
0
0
1 0
Sample Output
1
2
题目大意: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。
分析:我们可以用tarjan算法求出连通分量,我们可以把每个连通分量看为一个点,就成了一个新图,第一问求新图有多少个入度为0的个数(如果没有也要输出1)。第二问求的是max(入度个数,出度个数)。
const
MaxV=10000;
MaxE=10000;
type
arr=record
next,x,y:longint;
end;
var
n,m,tot,dindex,bcnt:longint;
ls,stack,low,dfn,belong,c,r:array[1..maxV]of longint;
g:array[1..MaxE]of arr;
instack:array[1..maxv] of boolean;
procedure init;
var
i,s,m:longint;
begin
readln(n);
m:=0;
fori:=1 to n do
begin
while 1=1 do
begin
read(s);
if s=0 then break;
inc(m);
with g[m] do
begin
x:=i; y:=s;
next:=ls[x];
ls[x]:=m;
end;
end;
end;
end;
function min(x,y:longint):longint;
begin
ifx>y then exit(y)
else exit(x);
end;
function max(x,y:longint):longint;
begin
ifx>y then exit(x)
else exit(y);
end;
procedure tarjan(i:longint);
var j,t:longint;
begin
inc(dindex);
low[i]:=dindex;
dfn[i]:=dindex;
inc(tot);
stack[tot]:=i;
t:=ls[i];
instack[i]:=true;
while t>0 do
begin
j:=g[t].y;
if dfn[j]=0 then
begin
tarjan(j);
low[i]:=min(low[i],low[j]);
end
else
if instack[j] then
low[i]:=min(low[i],dfn[j]);
t:=g[t].next;
end;
if low[i]=dfn[i] then
begin
inc(bcnt);
repeat
j:=stack[tot];
instack[j]:=false;
dec(tot);
belong[j]:=bcnt;
until i=j;
end;
end;
procedure solve;
var i:longint;
begin
dindex:=0;
bcnt:=0;
dindex:=0;
fillchar(dfn,sizeof(dfn),0);
for i:=1 to n do
if dfn[i]=0 then
begin
tarjan(i);
end;
end;
procedure main;
varj,i,k:longint;
begin
init;
solve;
for i:=1 to n do
begin
j:=ls[i];
while j<>0 do
begin
with g[j] do
begin
if belong[x]<>belong[y] then
begin
inc(c[belong[x]]);
inc(r[belong[y]]);
end;
j:=next;
end;
end;
end;
for i:=1 to bcnt do
if r[i]=0 then j:=j+1;
writeln(j);
k:=0;
for i:=1 to bcnt do
if c[i]=0 then inc(k);
if bcnt=1 then
begin
write(0);
halt;
end;
writeln(max(k,j));
end;
begin
while not seekeof do
begin
main;
end;
end.
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