bzoj 2039 最小割模型

　　比较明显的网络流最小割模型，对于这种模型我们需要先求获利的和，然后减去代价即可。

　　我们对于第i个人来说， 如果选他，会耗费A[I]的代价，那么(source,i,a[i])代表选他之后的代价，如果不选他，我们会产生Σw[i][j] 1<=j<=n的代价，也就是这么多的利益我们无法得到，然后对于两个人的互相影响，连边(i,j,2*w[i][j])，代表如果不选i，选j的话，本来i中选j的利益得不到，又要损失j对i的影响为w[i][j]。所以共计损失2*w[i][j]。之后求最小割=最大流就行了。

/**************************************************************
    Problem: 2039
    User: BLADEVIL
    Language: Pascal
    Result: Accepted
    Time: 860 ms
    Memory: 47112 kb
****************************************************************/

//By BLADEVIL
var
    n                           : longint ;
    pre, other, len             : array [ 0..4000010 ] of longint ;
    last                        : array [ 0..1010 ] of longint ;
    l                           : longint ;
    source, sink                : longint ;
    que, d                      : array [ 0..1010 ] of longint ;
    ans                         : longint ;

function min(a,b: longint ): longint ;
begin
    if a>b then min:=b else min:=a;
end ;

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 ;
    x                           : longint ;
    sum                         : longint ;
begin
    read(n);
    source:=n+ 2 ; sink:=source+ 1 ; l:= 1 ;
    for i:= 1 to n do
    begin
        read(x);
        connect(source,i,x);
        connect(i,source, 0 );
    end ;
    for i:= 1 to n do
    begin
        sum:= 0 ;
        for j:= 1 to n do
        begin
            read(x);
            if x= 0 then continue;
            sum:=sum+x;
            connect(i,j,x<< 1 );
            connect(j,i, 0 );
            ans:=ans+x;
        end ;
        connect(i,sink,sum);
        connect(sink,i, 0 );
    end ;
end ;

function bfs: boolean ;
var
    h, t, cur                   : longint ;
    q, p                        : longint ;
begin
    fillchar(d,sizeof(d), 0 );
    h:= 0 ; t:= 1 ;
    que[ 1 ]:=source;
    d[source]:= 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
    rest, tmp                   : longint ;
    q, p                        : 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(len[q],rest));
            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 div 10 );
    writeln (ans);
end ;

begin
    init;
    main;
end .