2245: [SDOI2011]工作安排

因为题目太长懒得看，所以现在才做这道题。

每加工一个零件就花费固定的费用，一看就是费用流。

建立S，T。  S向每个人连边，每一段怒气连一条流量为这一段零件数的边。 如果i可以加工j，add(i,j,inf,0)。每种零件向T连流量为个数的边。

zkw或spfa费用流都行。

因为w是递增的所以才能这么建边。

RunID	User	Problem	Result	Memory	Time	Language	Code_Length	Submit_Time
408065	lbz007	2245	Accepted	43300 kb	3188 ms	Pascal/Edit	2434 B	2013-05-08 17:19:07

const inf=1000000000;
var
  dis,tx,other,a,pre,next,head,e,x,long,w:array[0..500100]of int64;
  v:Array[0..100010]of boolean;
  can,kk,xx,tt,ee,ss,i,j,n,m:longint;
  sum,ans,mincost:int64;
function min(aa,bb:longint):longint;
  begin if aa>bb then exit(bb)  else exit(aa); end;
procedure add(u,v,c,cc:longint);
  begin
  inc(ee);x[ee]:=u;e[ee]:=v;long[ee]:=c;w[ee]:=cc;other[ee]:=ee+1;
  next[ee]:=head[u];head[u]:=ee;
  inc(ee);x[ee]:=v;e[ee]:=u;long[ee]:=0;w[ee]:=-cc;other[ee]:=ee-1;
  next[ee]:=head[v];head[v]:=ee;
  end;
function spfa :boolean;
  var i,h,t,j:longint;
  begin
  h:=1;t:=1;
  for i:=ss to tt do dis[i]:=inf;
  a[1]:=ss; pre[ss]:=-1;
  v[ss]:=true;  dis[ss]:=0;
  while h<=t do
    begin
    j:=head[a[h]];
    while j<>0 do
      begin
      if (long[j]>0)and(dis[a[h]]+w[j]<dis[e[j]]) then
        begin
        dis[e[j]]:=dis[a[h]]+w[j];
        pre[e[j]]:=j;
        if not v[e[j]] then
          begin  v[e[j]]:=true;  inc(t) ;a[t]:=e[j];  end;
        end;
      j:=next[j];
      end;
    v[a[h]]:=false;  inc(h);
    end;
  if dis[tt]>=inf-1000 then exit(false)
  else exit(true);
  end;  
procedure work;
  var zh,flow:longint;
  begin
  ans:=0;  mincost:=0;
  while spfa do
    begin
    zh:=tt;  flow:=maxlongint;
    while pre[zh]<>-1 do
      begin
      flow:=min(flow,long[pre[zh]]);
      zh:=x[pre[zh]];
      end;
    ans:=ans+flow;
    mincost:=mincost+dis[tt]*flow;
    zh:=tt;
    while pre[zh]<>-1 do
      begin
      dec(long[pre[zh]],flow);
      inc(long[other[pre[zh]]],flow);
      zh:=x[pre[zh]];
      end;
    end;
  end; 
    
begin
  readln(n,m);
  tt:=n+m+1;ss:=0;
  for i:=1 to m do
    begin
    read(xx);sum:=sum+xx;
    add(n+i,tt,xx,0);
    end;
  for i:=1 to n do
    begin
    for j:=1 to m do
      begin
      read(can);  
      if can=1 then
        add(i,n+j,inf,0);
      end;  
    readln;
    end;  
  for i:=1 to n do
    begin
    readln(kk);
    for j:=1 to kk do
      read(tx[j]);
    tx[kk+1]:=sum;  
    for j:=1 to kk+1 do
      begin
      read(xx);
      add(ss,i,tx[j]-tx[j-1],xx);
      end;
    end; 
  work;
  writeln(mincost);
  end.