【Tyvj1982】武器分配（费用流）

题意：有N个人要从A个物品中各取一个，B个物品中各取一个，选取第i个A类物品和第j个B类物品的费用是(a[i]-b[j])^2

求最小总花费

n<=a,b<=80 a[i],b[i]<=10000

思路：第一题费用流

由源点到每个A类物品连容量为1，费用为0的边

每个B类物品到第一个汇点连容量为1，费用为0的边

对于ai和bj连容量为1，费用为(a[i]-b[j])^2的边

因为只需要N对物品，由第一个汇点到第二个汇点连容量为N，费用为0的边来限制流量上限

答案就是从源点到第二个汇点流量为N的最小费用

跑SPFA费用流即可

PS：其实这道根据排序不等式是个DP，排序后做即可 但是谁叫你数据太小呢

 var q:array[..]of longint;
     head,vet,next,len1,len2,dis,fan:array[..]of longint;
     pre:array[..,..]of longint;
     c,d:array[..]of longint;
     inq:array[..]of boolean;
     n,m,i,source,src,tot,ans,a,b,j:longint;

 procedure add(a,b,c,d:longint);
 begin
  inc(tot);
  next[tot]:=head[a];
  vet[tot]:=b;
  len1[tot]:=c;
  len2[tot]:=d;
  head[a]:=tot;
  inc(tot);
  next[tot]:=head[b];
  vet[tot]:=a;
  len1[tot]:=;
  len2[tot]:=-d;
  head[b]:=tot;
 end;

 function min(x,y:longint):longint;
 begin
  if x<y then exit(x);
  exit(y);
 end;

 function spfa:boolean;
 var t,u,e,v,i,w:longint;
 begin
  for i:= to  do
  begin
   dis[i]:=maxlongint div ;
   inq[i]:=false;
  end;
  t:=-; w:=; q[]:=source; inq[source]:=true; dis[source]:=;
  while t<w do
  begin
   inc(t); u:=q[t mod ];
   inq[u]:=false;
   e:=head[u];
   while e<> do
   begin
    v:=vet[e];
    if (len1[e]>)and(dis[u]+len2[e]<dis[v]) then
    begin
     pre[v,]:=u;
     pre[v,]:=e;
     dis[v]:=dis[u]+len2[e];
     if not inq[v] then
     begin
      inc(w); q[w mod ]:=v;
      inq[v]:=true;
     end;
    end;
    e:=next[e];
   end;
  end;
  if dis[src]=maxlongint div  then exit(false)
   else exit(true);
 end;

 procedure mcf;
 var k,t,e:longint;
 begin
  k:=src; t:=maxlongint;
  while k<>source do
  begin
   t:=min(t,len1[pre[k,]]);
   k:=pre[k,];
  end;
  k:=src;
  while k<>source do
  begin
   e:=pre[k,];
   len1[e]:=len1[e]-t;
   len1[fan[e]]:=len1[fan[e]]+t;
   ans:=ans+t*len2[e];
   k:=pre[k,];
  end;
 end;

 begin
  assign(input,'tyvj1982.in'); reset(input);
  assign(output,'tyvj1982.out'); rewrite(output);
  readln(n,a,b);
  for i:= to  do
   if i mod = then fan[i]:=i+
    else fan[i]:=i-;
  for i:= to a do read(c[i]);
  for i:= to b do read(d[i]);
  for i:= to a do
   for j:= to b do add(i,a+j,,(c[i]-d[j])*(c[i]-d[j]));
  source:=; src:=;
  for i:= to a do add(source,i,,);
  for i:= to b do add(i+a,,,);
  add(,src,n,);
  while spfa do mcf;
  writeln(ans);
  close(input);
  close(output);
 end.