bzoj 3037 贪心

我们可以贪心的分析，每个点的入度如果是0，那么这个点不可能

被用来更新答案，那么我们每次找入度为0的点，将他去掉，如果他连的

点没有被更新过答案，那么更新答案，去掉该点，环的时候最后处理就行了

/**************************************************************

    Problem: 3037

    User: BLADEVIL

    Language: Pascal

    Result: Accepted

    Time:1672 ms

    Memory:12920 kb

****************************************************************/

 

//By BLADEVIL

var

    n, ans, cnt                 :longint;

    other, sum                  :array[0..1000020] of longint;

    flag                        :array[0..1000020] of boolean;

    que                         :array[0..1000020] of longint;

     

procedure init;

var

    i                           :longint;

begin

    readln(n);

    for i:=1 to n do

    begin

        read(other[i]);

        inc(sum[other[i]]);

    end;

end;

 

procedure main;

var

    h, t, cur                   :longint;

    i, j                        :longint;

     

begin

    t:=0;

    for i:=1 to n do

    if sum[i]=0 then

    begin

        inc(t);

        que[t]:=i;

    end;

    h:=1;

    while h<=t do

    begin

        cur:=que[h];

        inc(h);

        if (not flag[cur]) and (not flag[other[cur]]) then

        begin

            inc(ans);

            flag[other[cur]]:=true;

            dec(sum[other[other[cur]]]);

            if sum[other[other[cur]]]=0 then

            begin

                inc(t);

                que[t]:=other[other[cur]];

            end;

        end;

        flag[cur]:=true;

    end;

     

    for i:=1 to n do

        if  not flag[i] then

        begin

            cnt:=1;

            flag[i]:=true;

            j:=i;

            while other[j]<>i do

            begin

                flag[other[j]]:=true;

                inc(cnt);

                j:=other[j];

            end; 

            inc(ans,cnt div 2);

        end;

    writeln(ans);

end;

 

begin

    init;

    main;

end.