网络流建模

Posted 2020-08-22

poj3281 

有F种食物和D种饮料，每种食物或饮料只能供一头牛享用，且每头牛只享用一种食物和一种饮料。现在有N头牛，每头牛都有自己喜欢的食物种类列表和饮料种类列表，问最多能使几头牛同时享用到自己喜欢的食物和饮料。（1 <= F <= 100, 1 <= D <= 100, 1 <= N <= 100）

分析：

s和每种食物连边c=1

每种饮料和t连边c=1

然后牛拆成两个i1,i2放中间，喜欢的食物连i1，i2连喜欢的饮料c=1

代码：

program Dining;
var
  c:array[0..1000,0..1000]of longint;
  q:array[0..3000]of longint;
  sz:array[0..1000]of longint;
  n,i,m,num,m1,m2,x,v1,v2,j:longint;
function min(x,y:longint):longint;
begin
  if x<y then min:=x else min:=y;
end;
function bfs:boolean;
var h,t,x,i:longint;
begin
  fillchar(sz,sizeof(sz),0);
  h:=0; t:=1; q[1]:=0; sz[0]:=1;
  while h<t do
   begin
     inc(h); x:=q[h];
     for i:=0 to num do
      if (c[x,i]>0)and(sz[i]=0) then
       begin
         sz[i]:=sz[x]+1; inc(t); q[t]:=i;
       end;
   end;
  if sz[num]=0 then exit(false) else exit(true);
end;
function dfs(x,flow:longint):longint;
var d,i:longint;
begin
  if x=num then exit(flow);
  for i:=0 to num do
   if (sz[i]=sz[x]+1)and(c[x,i]>0) then
   begin
     d:=dfs(i,min(c[x,i],flow));
     if d<>0 then
      begin inc(c[i,x],d); dec(c[x,i],d); exit(d); end;
   end;
  exit(0);
end;
procedure solve;
var ans,s,flow,inf:longint;
begin
  ans:=0; s:=0; inf:=maxlongint div 3;
  while bfs do
   begin
     flow:=dfs(s,inf);
     while flow<>0 do
      begin
        inc(ans,flow); flow:=dfs(s,inf);
      end;
   end;
  writeln(ans);
end;
begin
  assign(input,‘Dining.in‘);
  reset(input);
  assign(output,‘Dining.out‘);
  rewrite(output);
  readln(n,m1,m2); num:=n*2+m1+m2+1;
  for i:=1 to m1 do c[0,i+n*2]:=1;
  for i:=1 to m2 do c[i+m1+n*2,num]:=1;
  for i:=1 to n do c[i,i+n]:=1;
  for i:=1 to n do
   begin
     read(v1,v2);
     for j:=1 to v1 do begin read(x); x:=x+n*2; c[x,i]:=1; end;
     for j:=1 to v2 do begin read(x); x:=x+n*2+m1; c[i+n,x]:=1; end;
     readln;
   end;
  solve;
  close(input); close(output);
end.     

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