bzoj 1001 平面图转对偶图 最短路求图最小割

原题传送门http://www.lydsy.com/JudgeOnline/problem.php?id=1001

整理了下之前A的题

平面图可以转化成对偶图，然后（NlogN）的可以求出图的最小割（最大流）

算法合集有具体的讲解，有兴趣的可以在网上搜下或者向我要（QQ30056882）

/**************************************************************
    Problem: 1001
    User: BLADEVIL
    Language: Pascal
    Result: Accepted
    Time:4124 ms
    Memory:158960 kb
****************************************************************/
 
//By BLADEVIL
var
    n, m                    :longint;
    pre, other, len, last   :array[0..6001000] of longint;
    l                       :longint;
    heng, shu, xie          :array[0..2010,0..2010] of longint;
    tot                     :longint;
    st, fin                 :longint;
    que, d                  :array[0..2000010] of longint;
    flag                    :array[0..2000010] of boolean;
     
procedure connect(x,y,z:longint);
begin
    inc(l);
    pre[l]:=last[x];
    last[x]:=l;
    other[l]:=y;
    len[l]:=z;
end;
     
procedure init;
var
    i, j                    :longint;
    min                     :longint;
begin
    read(n,m);
    for i:=1 to n do
        for j:=1 to m-1 do read(heng[i,j]);
     
    for i:=1 to n-1 do
        for j:=1 to m do read(shu[i,j]);
         
    for i:=1 to n-1 do
        for j:=1 to m-1 do read(xie[i,j]);
    min:=maxlongint;
    if (n=1) or (m=1) then
    begin
        for i:=1 to n do
            for j:=1 to m do
            begin
                if heng[i,j]>0 then if min>heng[i,j] then min:=heng[i,j];
                if shu[i,j]>0 then if min>shu[i,j] then min:=shu[i,j];
                if xie[i,j]>0 then if min>xie[i,j] then min:=xie[i,j];
            end;
        writeln(min);
        halt;
    end;
    tot:=2*(m-1)*(n-1);
    for i:=1 to tot do
    if i mod 2=0 then
    begin
        if ((i mod (2*(m-1)))>0) then
        begin
            connect(i,i+1,shu[i div (2*(m-1))+1,(i mod (2*(m-1))) div 2+1]);
            connect(i+1,i,shu[i div (2*(m-1))+1,(i mod (2*(m-1))) div 2+1]);
        end;
        if (i-2*m+1>0) then
            if ((i mod (2*(m-1)))>0) then
            begin
                connect(i,i-2*m+1,heng[i div (2*(m-1))+1,(i mod (2*(m-1))) div 2]); 
                connect(i-2*m+1,i,heng[i div (2*(m-1))+1,(i mod (2*(m-1))) div 2]);
            end else
            begin
                connect(i,i-2*m+1,heng[i div (2*(m-1)),m-1]);   
                connect(i-2*m+1,i,heng[i div (2*(m-1)),m-1]);
            end;
    end else
    begin
        connect(i,i+1,xie[i div (2*(m-1))+1,((i mod (2*(m-1)))+1) div 2]);
        connect(i+1,i,xie[i div (2*(m-1))+1,((i mod (2*(m-1)))+1) div 2]);
    end;
    st:=tot+1; fin:=tot+2;
    for i:=1 to (m-2) do
    begin
        connect(st,i*2,heng[1,i]);
        connect(i*2,st,heng[1,i]);
    end;
    for i:=2 to (n-1) do
    begin
        connect(st,i*2*(m-1),shu[i,m]);
        connect(i*2*(m-1),st,shu[i,m]);
    end;
    connect(st,2*(m-1),shu[1,m]);   
    connect(st,2*(m-1),heng[1,m-1]);    
    connect(2*(m-1),st,shu[1,m]);
    connect(2*(m-1),st,heng[1,m-1]);
    for i:=2 to (m-1) do
    begin
        connect((n-2)*2*(m-1)+2*i-1,fin,heng[n,i]);
        connect(fin,(n-2)*2*(m-1)+2*i-1,heng[n,i]);
    end;
    for i:=1 to (n-2) do
    begin
        connect(fin,(i-1)*2*(m-1)+1,shu[i,1]);
        connect((i-1)*2*(m-1)+1,fin,shu[i,1]);
    end;
    connect(fin,(n-2)*2*(m-1)+1,shu[(n-1),1]);
    connect(fin,(n-2)*2*(m-1)+1,heng[n,1]);
    connect((n-2)*2*(m-1)+1,fin,shu[(n-1),1]);
    connect((n-2)*2*(m-1)+1,fin,heng[n,1]);
end;
 
 
procedure main;
var
    i, j                    :longint;
    h, t                    :longint;
    cur                     :longint;
    q, p                    :longint;
     
begin
    que[1]:=st; 
    filldword(d,sizeof(d) div 4,maxlongint div 10);
    d[st]:=0;
    t:=1; h:=0;
    while h<>t do
    begin
        h:=h mod 2000000+1;
        cur:=que[h];
        flag[cur]:=false;
        q:=last[cur];
        while q<>0 do
        begin
            p:=other[q];
            if d[cur]+len[q]<d[p] then
            begin
                d[p]:=d[cur]+len[q];
                if not flag[cur] then
                begin
                    t:=t mod 2000000+1;
                    que[t]:=p;
                    flag[p]:=true;
                end;
            end;
            q:=pre[q];
        end;
    end;
    writeln(d[fin]);
end;    
 
begin
    ///assign(input,'stop.in'); reset(input);
    ///assign(output,'stop.out'); rewrite(output);
    init;
    main;
    ///close(output); close(output);
end.