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.