POJ3189 Steady Cow Assignment ——二分答案+二分图多重匹配——Pku3189

改进过后的匈牙利算法即可轻松秒掉最大流的各种NB算法。不过，这种方法有一个局限，就是右边集合可以匹配多个而左边集合只能匹配一个的时候才可以用，否则只能搞神马SAP等等了。

匈牙利算法的改进：

1、存储右边集合的结果时不要只存一个，而是将所有匹配结果都存下来。

2、在找增广路时，把（res[k]=0）的条件改成(res[k,0]<max[k])

CODE

Program Stead;//By_Thispoet

Const

	maxn=1000;

Var	

	pre,other,last							:Array[1..maxn*20]of Longint;

	res										:Array[1..20,0..maxn]of Longint;

	max										:Array[1..maxn]of Longint;

	state									:Array[1..20]of Boolean;

	rank									:Array[1..maxn,0..20]of Longint;

	h,t,i,j,k,n,m,l,r,mid,ans,sum,all		:Longint;



Function Dfs(i:Longint):Boolean;

var j,k,p:Longint;

begin

	j:=last[i];

	while j<>0 do

		begin

			k:=other[j];

			if not state[k] then

				begin

					state[k]:=true;

					if res[k,0]<max[k] then

						begin

							inc(res[k,0]);

							res[k,res[k,0]]:=i;

							exit(true);

						end else

							begin

								for p:=1 to res[k,0] do

									if Dfs(res[k,p]) then

										begin

											res[k,p]:=i;

											exit(true);

										end;

							end;

				end;

			j:=pre[j];

		end;

	exit(false);

end;





Function Check(pos:Longint):Boolean;

begin

	if m<=pos then exit(true);

	for h:=1 to m-pos do

		begin

			k:=0;

			fillchar(last,sizeof(last),0);

			for i:=1 to n do

				for j:=h to h+pos do

					begin

						inc(k);pre[k]:=last[i];

						last[i]:=k;other[k]:=rank[i,j];

					end;

			sum:=0;

			for i:=1 to m do res[i,0]:=0;

			for i:=1 to n do

				begin

					fillchar(state,sizeof(state),0);

					if Dfs(i) then inc(sum) else break;

				end;

			if sum=n then exit(true);

		end;

	exit(false);

end;





BEGIN





	readln(n,m);

	for i:=1 to n do

		for j:=1 to m do read(rank[i,j]);

	for i:=1 to m do read(max[i]);

	l:=0;r:=20;

	all:=0;

	while l<=r do

		begin

			mid:=(l+r)>>1;

			if check(mid) then

				begin

					ans:=mid;

					r:=mid-1;

				end else l:=mid+1;

		end;

	

	writeln(ans+1);

	

	

END.