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| program jsp;
const nn=20;
var f:array[1..nn,0..nn*nn]of boolean;//each machine condition;
t:array[1..nn,1..nn]of integer;//cover time;
e:array[1..nn,0..nn]of integer;//end time;
a:array[1..nn*nn]of integer;//following;
b:array[1..nn,1..nn]of integer;//belong to machine
g:array[1..nn*nn]of integer;//dealing with;
flag:boolean;
i,j,m,n,k,best,nt,nm:integer;
begin
assign(input,'jsp.in');reset(input);
assign(output,'');rewrite(output);
readln(m,n);
fillchar(a,sizeof(a),0);
fillchar(e,sizeof(e),0);
fillchar(f,sizeof(f),true);
fillchar(g,sizeof(g),0);
for i:=1 to m*n do read(a[i]);readln;
for i:=1 to n do begin
for j:=1 to m do
read(b[i,j]);
end;
for i:=1 to n do begin
for j:=1 to m do
read(t[i,j]);
readln;
end;
for i:=1 to m*n do begin
inc(g[a[i]]);
nt:=t[a[i],g[a[i]]];
nm:=b[a[i],g[a[i]]];{nt：needtime；nm：machine}
for j:=e[a[i],g[a[i]]-1]to 400 do begin{从该工件的上一个工序结束时间开始搜索}
flag:=true;
for k:=j to j+nt do
if k<=400 then
flag:=flag and f[nm,k];
if flag then begin{如果搜索成功}
e[a[i],g[a[i]]]:=j+nt;{记下结束时间}
for k:=j+1 to j+nt-1 do f[nm,k]:=false;
break;{减少搜索}
end;
end;
end;
best:=0;
for i:=1 to n do {从每道工序入手，但如果从每个机器入手呢？}
if best<e[i,m]then best:=e[i,m];
writeln(best);
close(input);close(output);
end.
|
