bzoj 1412 最小割 网络流

　　比较明显的最小割建模， 因为我们需要把狼和羊分开。

　　那么我们连接source和每个羊，流量为inf，代表这条边不能成为最小割中的点，同理连接每个狼和汇，流量为inf，正确性同上，那么对于每个相邻的羊和狼，连接边，流量为1，代表隔开这两个点需要1的代价，对于每个空地和狼或者羊，连接边，流量为1，代表隔开这个两个点的代价为1，同时需要注意的是，对于空地之间的连边也应该是1，因为很有可能狼和羊通过空地相遇。这样做最大流就行了。

　　反思：手残将空地之间的连成inf了。。。

/**************************************************************
    Problem: 1412
    User: BLADEVIL
    Language: Pascal
    Result: Accepted
    Time:112 ms
    Memory:3152 kb
****************************************************************/
 
//By BLADEVIL
var
    n, m                        :longint;
    num, map                    :array[0..210,0..210] of longint;
    pre, other, len             :array[0..200010] of longint;
    last                        :array[0..20010] of longint;
    l                           :longint;
    source, sink                :longint;
    go                          :array[0..2,0..4] of longint;
    que, d                      :array[0..20010] of longint;
    ans                         :longint;
     
procedure connect(x,y,z:longint);
begin
    inc(l);
    pre[l]:=last[x];
    last[x]:=l;
    other[l]:=y;
    len[l]:=z;
end;
     
function min(a,b:longint):longint;
begin
    if a>b then exit(b) else exit(a);
end;
     
procedure init;
var
    i, j, k                     :longint;
    x                           :longint;
    nx, ny                      :longint;
begin
    go[1,1]:=-1; go[2,2]:=1; go[1,3]:=1; go[2,4]:=-1;
    read(n,m);
    l:=1;
    for i:=1 to n do
        for j:=1 to m do num[i,j]:=(i-1)*m+j;
     
    source:=n*m+2; sink:=source+1;
     
    for i:=1 to n do
        for j:=1 to m do read(map[i,j]);
     
    for i:=1 to n do
        for j:=1 to m do
        begin
            if map[i,j]=1 then
            begin
                connect(source,num[i,j],maxlongint);
                connect(num[i,j],source,0);
            end else
            if map[i,j]=2 then
            begin
                connect(num[i,j],sink,maxlongint);
                connect(sink,num[i,j],0);
            end;
            for k:=1 to 4 do
            begin
                nx:=i+go[1,k];
                ny:=j+go[2,k];
                if (nx<1) or (nx>n) or (ny<1) or (ny>m) then  continue;
                if map[i,j]<>map[nx,ny] then
                begin
                    connect(num[i,j],num[nx,ny],1);
                    connect(num[nx,ny],num[i,j],0);
                end;
                if (map[i,j]=map[nx,ny]) and (map[i,j]=0) then
                begin
                    connect(num[i,j],num[nx,ny],1);
                    connect(num[nx,ny],num[i,j],0);
                end;
            end;
        end;
end;
 
function bfs:boolean;
var
    q, p                        :longint;
    h, t, cur                   :longint;
begin
    fillchar(d,sizeof(d),0);
    que[1]:=source;
    d[source]:=1;
    h:=0; t:=1;
    while h<t do
    begin
        inc(h);
        cur:=que[h];
        q:=last[cur];
        while q<>0 do
        begin
            p:=other[q];
            if (len[q]>0) and (d[p]=0) then
            begin
                inc(t);
                que[t]:=p;
                d[p]:=d[cur]+1;
                if p=sink then exit(true);
            end;
            q:=pre[q];
        end;
    end;
    exit(false);
end;
 
function dinic(x,flow:longint):longint;
var
    q, p                        :longint;
    tmp, rest                   :longint;
begin
    if x=sink then exit(flow);
    rest:=flow;
    q:=last[x];
    while q<>0 do
    begin
        p:=other[q];
        if (len[q]>0) and (d[p]=d[x]+1) and (rest>0) then
        begin
            tmp:=dinic(p,min(rest,len[q]));
            dec(rest,tmp);
            dec(len[q],tmp);
            inc(len[q xor 1],tmp);
        end;
        q:=pre[q];
    end;
    exit(flow-rest);
end;
 
procedure main;
begin
    while bfs do
        ans:=ans+dinic(source,maxlongint);
    writeln(ans);
end;
 
begin
    init;
    main;
end.