codevs天梯棋盘染色2

来源：互联网发布：电脑里编程开发编辑：程序博客网时间：2024/05/06 05:21

有一个5*N的棋盘，棋盘中的一些格子已经被染成了黑色，你的任务是对最少的格子染色，使得所有的黑色能连成一块。

分析:这题很容易想到是广搜，但数据那么大，如果一个一个点来更新就gg了==。所以开出一个树状数组对最大块进行预处理，剩下就没什么好说的了。

const
maxn=102;
type
node=record
x,y:longint;
end;
aa=array[0..6]of longint;
var
f:array[0..maxn,0..1204]of longint;
a:array[0..maxn]of longint;
fa:array[1..4]of longint;
flag:array[1..4]of boolean;
n:longint;
procedure init;
var
i,j:longint;
s:char;
begin
readln(n);
for i:=1 to n do
begin
for j:=1 to 5 do
begin
read(s);
a[i]:=a[i]<<1+ord(s)-ord('0');
end;
readln;
end;
while n>0 do
begin
if a[n]>0 then break;
dec(n);
end;
if n=0 then
begin
write(0);
halt;
end;
end;
procedure change(var a:aa;b,c:longint);
var
i:longint;
begin
for i:=1 to 5 do
if a[i]=b then
begin
a[i]:=c;
if (a[i-1]<>0) and (a[i-1]<10) then change(a,a[i-1],c);
if (a[i+1]<>0) and (a[i+1]<10) then change(a,a[i+1],c);
end;
end;
procedure get(var a:aa);
var
i,c:longint;
begin
c:=10;
for i:=1 to 5 do
if (a[i]<>0) and (a[i]<10) then
begin
inc(c);
change(a,a[i],c);
end;
for i:=1 to 5 do
if a[i]>0 then dec(a[i],10);
end;
function bit(x:longint):longint;
begin
if x=0 then exit(0);
exit(bit(x-(x and -x))+1);
end;
var
q:array[0..maxn*1024]of node;
procedure work;
var
head,tail,i,j,k,ans,save:longint;
s,t:aa;
flag:boolean;
begin
fillchar(f,sizeof(f),1);
t[0]:=0;
t[6]:=0;
ans:=500;
f[0,0]:=0;
q[1].x:=0;
q[1].y:=0;
head:=1;
tail:=1;
while head<=tail do
begin
save:=q[head].y;
for i:=1 to 5 do
begin
s[i]:=q[head].y and 3;
q[head].y:=q[head].y>>2;
end;
q[head].y:=save;
if q[head].x=n then
begin
flag:=true;
for i:=1 to 5 do
if s[i]>1 then flag:=false;
if flag then
if ans>f[q[head].x,q[head].y] then ans:=f[q[head].x,q[head].y];
inc(head);
continue;
end;
for i:=0 to 31 do
if i and a[q[head].x+1]=0 then
begin
for j:=1 to 5 do
t[j]:=(((a[q[head].x+1]+i)>>(j-1))and 1)*(j+3);
k:=0;
for j:=1 to 5 do
if (s[j]>0) and (t[j]>0) then
begin
t[j]:=s[j];
k:=k or (1<<s[j]);
end;
flag:=true;
for j:=1 to 5 do
if (s[j]>0) and (k and (1<<s[j])=0) then flag:=false;
if flag=false then continue;
get(t);
k:=0;
for j:=5 downto 1 do
k:=k<<2+t[j];
if f[q[head].x+1,k]>500 then
begin
inc(tail);
q[tail].x:=q[head].x+1;
q[tail].y:=k;
end;
if f[q[head].x+1,k]>f[q[head].x,q[head].y]+bit(i) then f[q[head].x+1,k]:=f[q[head].x,q[head].y]+bit(i);
end;
inc(head);
end;
write(ans);
end;
begin
init;
work;
end.

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codevs天梯棋盘染色2