自動車給油走行問題(car)
2238 ワード
「アルゴリズム設計と実験問題解」という本を強く軽蔑し、本の問題解は根本的に別の問題であり、答えには対応できない.
アルゴリズム:DP
分析:work[i,j,k]で(i,j)という点の残油量kの最小費用を表す.
アルゴリズム:DP
分析:work[i,j,k]で(i,j)という点の残油量kの最小費用を表す.
program car;
const
maxn=100;
maxoil=12;
var
n,go,cost,gocost,build:longint;
map:array [0..maxn,0..maxn] of longint;{map[i,j] 。}
work:array [0..maxn,0..maxn,0..maxoil] of longint;{}
s:array [0..4,0..3] of longint;{s , , 。}
procedure init;
var
i,j:longint;
begin
readln(n,go,cost,gocost,build);
for i:=0 to n-1 do
begin
for j:=0 to n-1 do read(map[i,j]);
readln;
end;
s[0,0]:=-1;
s[0,1]:=0;
s[0,2]:=0;
s[1,0]:=0;
s[1,1]:=-1;
s[1,2]:=0;
s[2,0]:=1;
s[2,1]:=0;
s[2,2]:=gocost;
s[3,0]:=0;
s[3,1]:=1;
s[3,2]:=gocost;
end;
procedure main;
var
i,y,j,p,q,min,x:longint;
begin
fillchar(work,sizeof(work),100);{ 。}
for i:=0 to go do work[0,0,i]:=0;
y:=1;
while y<>0 do
begin
y:=0;
for i:=0 to n-1 do
begin
for j:=0 to n-1 do
begin
if (i<>0) or (j<>0) then{ 。}
begin
for p:=0 to go do{ 。}
begin
min:=1000000;{ 。}
for q:=0 to 3 do
begin
{ 。}
if (i=0) and (q=0) then continue;
if (j=0) and (q=1) then continue;
if (i=n-1) and (q=2) then continue;
if (j=n-1) and (q=3) then continue;
if work[i+s[q,0],j+s[q,1],p+1]+s[q,2]min+cost*map[i,j] then inc(y);
work[i,j,p]:=min;
if map[i,j]=1 then
begin
inc(work[i,j,0],cost);
for x:=1 to go do work[i,j,x]:=work[i,j,0];
break;
end;
end
else
begin
work[i,j,p]:=work[i,j,0]+cost+build;
for x:=p+1 to go do work[i,j,x]:=work[i,j,p];
break;
end;
end;
end;
end;
end;
end;
end;
begin
assign(input,'car.in'); reset(input);
assign(output,'car.out'); rewrite(output);
init;
main;
writeln(work[n-1,n-1,0]);
close(input); close(output);
end.