bzoj 1834

网络流的模板题

首先第一问我们直接用dinic搞就行了，费用直接存为0（时间上界非常松，这道题是能过），然后第二问我们只需要在第一问

的残余网络上加一个源点，源点指向1号点，容量为k，费用为0，然后对于之前的每一条边，建一个相同的边

容量为无穷大，费用为原来的费用。

//By BLADEVIL

var

    pre, other, len, w            :array[0..20100] of longint;

    last                        :array[0..2100] of longint;

    l                            :longint;

    n, m, k                     :longint;

    father, que, dis            :array[0..2100] of longint;

    flag                        :array[0..2100] of boolean;

    a, b, c, d                    :array[0..20100] of longint;

    cost                        :longint;

    

function min(a,b:longint):longint;

begin

    if a>b then min:=b else min:=a;

end;

    

procedure connect(a,b,c,d:longint);

begin

    inc(l);

    pre[l]:=last[a];

    last[a]:=l;

    other[l]:=b;

    len[l]:=c;

    w[l]:=d;

end;



procedure init;

var

    i                            :longint;

    

begin

    read(n,m,k);

    l:=1;

    for i:=1 to m do read(a[i],b[i],c[i],d[i]);

    for i:=1 to m do 

    begin

        connect(a[i],b[i],c[i],0);

        connect(b[i],a[i],0,0);

    end;

end;



function bfs:boolean;

var

    q, p                        :longint;

    cur                            :longint;

    h, t                        :longint;

begin

    fillchar(dis,sizeof(dis),0);

    h:=0; t:=1;

    que[1]:=1; dis[1]:=1;

    while t<>h do 

    begin

        inc(h);

        cur:=que[h];

        q:=last[cur];

        while q<>0 do 

        begin

            p:=other[q];

            if (len[q]>0) and (dis[p]=0) then 

            begin

                inc(t);

                que[t]:=p;

                dis[p]:=dis[cur]+1;

                if p=n then exit(true);

            end;

            q:=pre[q];

        end;

    end;

    exit(false);

end;



function dinic(x,flow:longint):longint;

var

    q, p                        :longint;

    rest, tmp                    :longint;

begin

    if x=n then exit(flow);

    rest:=flow;

    q:=last[x];

    while q<>0 do

    begin

        p:=other[q];

        if (len[q]>0) and (rest>0) and (dis[p]=dis[x]+1) 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 spfa;

var

    q, p                        :longint;

    h, t, cur                    :longint;

    

begin

    fillchar(flag,sizeof(flag),false);

    filldword(dis,sizeof(dis) div 4,maxlongint div 10);

    h:=0; t:=1;

    que[1]:=n+1;

    dis[n+1]:=0;

    while h<>t do 

    begin

        h:=h mod 2000+1;

        cur:=que[h];

        flag[cur]:=false;

        q:=last[cur];

        while q<>0 do

        begin

            if len[q]>0 then 

            begin

                p:=other[q];

                if dis[cur]+w[q]<dis[p] then

                begin

                    dis[p]:=dis[cur]+w[q];

                    father[p]:=q;

                    if not flag[p] then 

                    begin

                        t:=t mod 2000+1;

                        que[t]:=p;

                        flag[p]:=true;

                    end;

                end;

            end;

            q:=pre[q];

        end;

    end;

end;



procedure update;

var

    cur                            :longint;

    low                            :longint;

begin

    cur:=n;

    low:=maxlongint div 10;

    while cur<>n+1 do 

    begin

        low:=min(low,len[father[cur]]);

        cur:=other[father[cur] xor 1];

    end;

    cur:=n;

    while cur<>n+1 do 

    begin

        inc(cost,low*w[father[cur]]);

        dec(len[father[cur]],low);

        inc(len[father[cur] xor 1],low);

        cur:=other[father[cur] xor 1];

    end;

end;

    

procedure main;

var

    ans                            :longint;

    i                            :longint;

    

begin

    ans:=0;

    while bfs do 

        ans:=ans+dinic(1,maxlongint div 10);



    for i:=1 to m do 

    begin

        connect(a[i],b[i],maxlongint div 10,d[i]);

        connect(b[i],a[i],0,-d[i]);

    end;

    

    connect(n+1,1,k,0);

    connect(1,n+1,0,0);

    while true do

    begin

        spfa;

        if dis[n]=maxlongint div 10 then break;

        update;

    end;

    writeln(ans,' ',cost);

end;



begin

    init;

    main;

end.