【BZOJ2127】happiness（最小割）

题意：高一一班的座位表是个n*m的矩阵，经过一个学期的相处，每个同学和前后左右相邻的同学互相成为了好朋友。

这学期要分文理科了，每个同学对于选择文科与理科有着自己的喜悦值，

而一对好朋友如果能同时选文科或者理科，那么他们又将收获一些喜悦值。

作为计算机竞赛教练的scp大老板，想知道如何分配可以使得全班的喜悦值总和最大。

对于100%以内的数据，n,m<=100 所有喜悦值均为小于等于5000的非负整数

思路：同BZOJ3894

双倍经验

话说这是一个经典网络流模型

var head,vet,next,len,gap,dis,fan:array[0..1100000]of longint;
    a,num:array[1..100,1..100,1..2]of longint;
    n,m,tot,i,j,sum,s,source,src,k,x,y,tmp:longint;

procedure add(a,b,c:longint);
begin
 inc(tot);
 next[tot]:=head[a];
 vet[tot]:=b;
 len[tot]:=c;
 head[a]:=tot;

 inc(tot);
 next[tot]:=head[b];
 vet[tot]:=a;
 len[tot]:=0;
 head[b]:=tot;
end;

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

function dfs(u,aug:longint):longint;
var e,v,val,flow,t:longint;
begin
 if u=src then exit(aug);
 e:=head[u]; val:=s-1; flow:=0;
 while e<>0 do
 begin
  v:=vet[e];
  if len[e]>0 then
  begin
   if dis[u]=dis[v]+1 then
   begin
    t:=dfs(v,min(len[e],aug-flow));
    len[e]:=len[e]-t;
    len[fan[e]]:=len[fan[e]]+t;
    flow:=flow+t;
    if dis[source]>=s then exit(flow);
    if aug=flow then break;
   end;
   val:=min(val,dis[v]);
  end;
  e:=next[e];
 end;
 if flow=0 then
 begin
  dec(gap[dis[u]]);
  if gap[dis[u]]=0 then dis[source]:=s;
  dis[u]:=val+1;
  inc(gap[dis[u]]);
 end;
 exit(flow);
end;

function maxflow:longint;
var ans:longint;
begin
 fillchar(gap,sizeof(gap),0);
 fillchar(dis,sizeof(dis),0);
 gap[0]:=s; ans:=0;
 while dis[source]<s do ans:=ans+dfs(source,maxlongint);
 exit(ans);
end;

begin
 assign(input,'bzoj2127.in'); reset(input);
 assign(output,'bzoj2127.out'); rewrite(output);
 readln(n,m);
 for i:=1 to 1100000 do
  if i and 1=1 then fan[i]:=i+1
   else fan[i]:=i-1;
 for i:=1 to 2 do
  for j:=1 to n do
   for k:=1 to m do
   begin
    read(a[j,k,i]);
    inc(s); num[j,k,i]:=s;
    sum:=sum+a[j,k,i];
   end;
 tmp:=s;
 s:=2*(n*m+(n-1)*m+(m-1)*n);
 source:=s+1; src:=s+2;
 s:=s+2;
 for i:=1 to n do
  for j:=1 to m do add(source,num[i,j,1],a[i,j,1]);
 for i:=1 to n do
  for j:=1 to m do add(num[i,j,2],src,a[i,j,2]);
 for i:=1 to n do
  for j:=1 to m do add(num[i,j,1],num[i,j,2],maxlongint);
 for i:=1 to n-1 do
  for j:=1 to m do
  begin
   read(x); inc(tmp); sum:=sum+x;
   add(source,tmp,x);
   add(tmp,num[i,j,1],maxlongint);
   add(tmp,num[i+1,j,1],maxlongint);
  end;
 for i:=1 to n-1 do
  for j:=1 to m do
  begin
   read(x); inc(tmp); sum:=sum+x;
   add(tmp,src,x);
   add(num[i,j,2],tmp,maxlongint);
   add(num[i+1,j,2],tmp,maxlongint);
  end;
 for i:=1 to n do
  for j:=1 to m-1 do
  begin
   read(x); inc(tmp); sum:=sum+x;
   add(source,tmp,x);
   add(tmp,num[i,j,1],maxlongint);
   add(tmp,num[i,j+1,1],maxlongint);
  end;
 for i:=1 to n do
  for j:=1 to m-1 do
  begin
   read(x); inc(tmp); sum:=sum+x;
   add(tmp,src,x);
   add(num[i,j,2],tmp,maxlongint);
   add(num[i,j+1,2],tmp,maxlongint);
  end;

 writeln(sum-maxflow);
end.




 close(input);
 close(output);
end.