poj2195 Going Home

这题可以看出是一个裸的km算法题，直接套模板即可，同时要注意的是在算的时候先取反.

（思路二：其实还可以转化为求最小费用，设一个超级源点把人连起来，各超级汇点把房间连起来，然后求解即可）

var
n,m,k1,k2,delta:longint;
w:array[1..100,1..100]of longint;
lx,ly,match:array[1..100]of longint;
a,b:array[1..100,1..2]of longint;
hx,hy,h:array[1..100]of boolean;
procedure init;
var i,j:longint;c:char;
begin
k1:=0;k2:=0;
fillchar(lx,sizeof(lx),0);
fillchar(ly,sizeof(ly),0);
fillchar(match,sizeof(match),0);
for i:=1 to n do
begin
for j:=1 to m do
begin
read(c);
if(c='m')then begin inc(k1);a[k1][1]:=i;a[k1][2]:=j; end;
if(c='H')then begin inc(k2);b[k2][1]:=i;b[k2][2]:=j; end;
end;
readln;
end;
for i:=1 to k1 do
for j:=1 to k2 do
w[i,
4000
j]:=-(abs(a[i][1]-b[j][1])+abs(a[i][2]-b[j][2]));
end;
function max(a,b:longint):longint;
begin
if(a>b)then exit(a) else exit(b);
end;
function check(k:longint):boolean;
var i,t:longint;
begin
hx[k]:=true;
for i:=1 to k2 do
if(not hy[i])then
begin
t:=lx[k]+ly[i]-w[k,i];
if(t=0)then
begin
hy[i]:=true;
if(match[i]=0)or(check(match[i]))then
begin
match[i]:=k;
exit(true);
end;
end
else begin
if(delta>t)then delta:=t;
end;
end;
exit(false);
end;
procedure km;
var i,j,ans:longint;
begin
for i:=1 to k1 do
for j:=1 to k2 do
lx[i]:=max(lx[i],w[i,j]);
for i:=1 to k1 do
begin
while(true)do
begin
fillchar(hx,sizeof(hx),0);
fillchar(hy,sizeof(hy),0);
delta:=maxlongint>>2;
if(check(i))then break;
for j:=1 to k1 do
if(hx[j])then dec(lx[j],delta);
for j:=1 to k2 do
if(hy[j])then inc(ly[j],delta);
end;
end;
ans:=0;
for i:=1 to k2 do
ans:=ans+lx[match[i]]+ly[i];
writeln(-ans);
end;
begin
readln(n,m);
while(n<>0)do
begin
init;
km;
readln(n,m);
end;
end.