【BZOJ1412】狼和羊的故事（最小割）

题意：将一个由0,1,2构成的矩阵里的1与2全部分割最少需要选取多少条边

n,m<=100 

 

思路：裸的最小割模型

相邻的格子连容量为1的边（其实可以少连很多遍，1与1，2与2之间的边是没有意义的）

由源点到所有1连容量为oo的边，2到汇点连容量为oo的边

最小割即是答案

  1 const dx:array[1..4]of longint=(-1,0,0,1);
  2       dy:array[1..4]of longint=(0,-1,1,0);
  3 var head,vet,next,len,dis,gap,fan:array[0..200000]of longint;
  4     num,a:array[1..300,1..300]of longint;
  5     n,m,i,j,x,y,tot,s,source,src,k:longint;
  6 
  7 procedure add(a,b,c:longint);
  8 begin
  9  inc(tot);
 10  next[tot]:=head[a];
 11  vet[tot]:=b;
 12  len[tot]:=c;
 13  head[a]:=tot;
 14 end;
 15 
 16 function min(x,y:longint):longint;
 17 begin
 18  if x<y then exit(x);
 19  exit(y);
 20 end;
 21 
 22 function dfs(u,aug:longint):longint;
 23 var e,v,val,t,flow:longint;
 24 begin
 25  if u=src then exit(aug);
 26  e:=head[u]; val:=s-1; flow:=0;
 27  while e<>0 do
 28  begin
 29   v:=vet[e];
 30   if len[e]>0 then
 31   begin
 32    if dis[u]=dis[v]+1 then
 33    begin
 34     t:=dfs(v,min(aug-flow,len[e]));
 35     len[e]:=len[e]-t;
 36     len[fan[e]]:=len[fan[e]]+t;
 37     flow:=flow+t;
 38     if dis[source]>=s then exit(flow);
 39     if aug=flow then break;
 40    end;
 41    val:=min(val,dis[v]);
 42   end;
 43   e:=next[e];
 44  end;
 45  if flow=0 then
 46  begin
 47   dec(gap[dis[u]]);
 48   if gap[dis[u]]=0 then dis[source]:=s;
 49   dis[u]:=val+1;
 50   inc(gap[dis[u]]);
 51  end;
 52  exit(flow);
 53 end;
 54 
 55 function maxflow:longint;
 56 var ans:longint;
 57 begin
 58  fillchar(gap,sizeof(gap),0);
 59  fillchar(dis,sizeof(dis),0);
 60  gap[0]:=s; ans:=0;
 61  while dis[source]<s do ans:=ans+dfs(source,maxlongint);
 62  exit(ans);
 63 end;
 64 
 65 begin
 66  assign(input,'bzoj1412.in'); reset(input);
 67  assign(output,'bzoj1412.out'); rewrite(output);
 68  readln(n,m);
 69  for i:=1 to n do
 70   for j:=1 to m do
 71   begin
 72    read(a[i,j]);
 73    inc(s); num[i,j]:=s;
 74   end;
 75  for i:=1 to 200000 do
 76   if i mod 2=1 then fan[i]:=i+1
 77    else fan[i]:=i-1;
 78  for i:=1 to n do
 79   for j:=1 to m do
 80    for k:=1 to 4 do
 81    begin
 82     x:=i+dx[k]; y:=j+dy[k];
 83     if (x>0)and(x<=n)and(y>0)and(y<=m) then
 84     begin
 85      add(num[i,j],num[x,y],1);
 86      add(num[x,y],num[i,j],0);
 87     end;
 88    end;
 89  source:=s+1; src:=s+2; s:=s+2;
 90  for i:=1 to n do
 91   for j:=1 to m do
 92   begin
 93    if a[i,j]=1 then
 94    begin
 95     add(source,num[i,j],maxlongint);
 96     add(num[i,j],source,0);
 97    end;
 98    if a[i,j]=2 then
 99    begin
100     add(num[i,j],src,maxlongint);
101     add(src,num[i,j],0);
102    end;
103   end;
104 
105  writeln(maxflow);
106  close(input);
107  close(output);
108 end.

 

posted on 2016-12-16 20:28  myx12345  阅读(231)  评论(0)  编辑  收藏  举报

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