bzoj 3232 01分数规划+最大权封闭子图判定

我们的目标是使v/c最小化，所以构造函数g(x)=v-x*c，那么

二分一个X，判断当时的v-x*c的值是多少，然后根据g(x)函数的

单调递减性来二分，判断，直到g(x)=0的时候当前的X就是答案。

然后我直接写的tle了，这是这两天tle的第3道题了。。。再改改。。。

/**************************************************************
    Problem: 3232
    User: BLADEVIL
    Language: Pascal
    Result: Time_Limit_Exceed
****************************************************************/
 
//By BLADEVIL
const
    lim                         =1e-5;
     
var
    n, m                        :longint;
    pre, other                  :array[0..100010] of longint;
    len                         :array[0..100010] of extended;
    last                        :array[0..3010] of longint;
    tot                         :longint;
    num                         :array[0..60,0..60] of longint;
    key, heng, shu              :array[0..60,0..60] of longint;
    sum                         :longint;
    print                       :extended;
    que, d                      :array[0..3010] of longint;
    source, sink                :longint;
     
function min(a,b:extended):extended;
begin
    if a>b then min:=b else min:=a;
end;
 
function judge(x:extended):extended;
begin
    if abs(x)<lim then exit(0);
    if x<0 then exit(-1) else exit(1);
end;
     
procedure connect(x,y:longint;z:extended);
begin
    inc(tot);
    pre[tot]:=last[x];
    last[x]:=tot;
    other[tot]:=y;
    len[tot]:=z;
end;
     
procedure init;
var
    i, j                        :longint;
     
begin
    read(n,m);
    for i:=1 to n do
        for j:=1 to m do num[i,j]:=(i-1)*m+j;
    for i:=1 to n do
        for j:=1 to m do
        begin
            read(key[i,j]);
            sum:=sum+key[i,j];
        end;
    for i:=1 to n+1 do
        for j:=1 to m do read(heng[i,j]);
    for i:=1 to n do
        for j:=1 to m+1 do read(shu[i,j]);
    source:=num[n,m]+2;
    sink:=source+1;
end;
 
function bfs:boolean;
var
    q, p                        :longint;
    h, t, cur                   :longint;
begin
    fillchar(d,sizeof(d),0);
    d[source]:=1;
    h:=0; t:=1; que[1]:=source;
    while h<t do
    begin
        inc(h);
        cur:=que[h];
        q:=last[cur];
        while q<>0 do
        begin
            p:=other[q];
            if (judge(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:longint;flow:extended):extended;
var
    rest, tmp                   :extended;
    q, p                        :longint;
     
begin
    if x=sink then exit(flow);
    rest:=flow;
    q:=last[x];
    while q<>0 do
    begin
        p:=other[q];
        if (judge(len[q])>0) and (d[p]=d[x]+1) and (rest>0) then
        begin
            tmp:=dinic(p,min(rest,len[q]));
            rest:=rest-tmp;
            len[q]:=len[q]-tmp;
            len[q xor 1]:=len[q xor 1]+tmp;
        end;
        q:=pre[q];
    end;
    exit(flow-rest);
end;
 
procedure main;
var
    l, r, mid                   :extended;
    cur                         :longint;
    ans                         :extended;
    i, j                        :longint;
     
begin
    l:=0; r:=90;
    while r-l>lim do
    begin
        mid:=(l+r)/2;
        fillchar(last,sizeof(last),0);
        tot:=1;
        for i:=1 to n do
            for j:=1 to m do
            begin
                connect(source,num[i,j],key[i,j]);
                connect(num[i,j],source,0);
            end;
         
        for i:=1 to n do
            for j:=1 to m do
            begin
                cur:=0;
                if i=1 then inc(cur,heng[i,j]);
                if i=n then inc(cur,heng[i+1,j]);
                if j=1 then inc(cur,shu[i,j]);
                if j=m then inc(cur,shu[i,j+1]);
                if cur>0 then
                begin
                    connect(num[i,j],sink,cur*mid);
                    connect(sink,num[i,j],0);
                end;
            end;
        for i:=1 to n-1 do
            for j:=1 to m do
            begin
                connect(num[i,j],num[i+1,j],heng[i+1,j]*mid);
                connect(num[i+1,j],num[i,j],heng[i+1,j]*mid);
            end;
        for i:=1 to n do
            for j:=1 to m-1 do
            begin
                connect(num[i,j],num[i,j+1],shu[i,j+1]*mid);
                connect(num[i,j+1],num[i,j],shu[i,j+1]*mid);
            end;
        ans:=0;
        while bfs do
            ans:=ans+dinic(source,maxlongint);
        if judge(sum-ans)>0 then l:=mid else r:=mid;
    end;
    writeln(l:0:3);
end;
 
begin
    init;
    main;
end.