poj2195 Going Home

這題可以看出是一個裸的km算法題，直接套模板即可，同時要注意的是在算的時候先取反.

（思路二：其實還可以轉化爲求最小費用，設一個超級源點把人連起來，各超級匯點把房間連起來，然後求解即可）

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,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.