• bzoj 1051 tarjan强连通分量


    2013-11-16 11:39

    原题传送门http://www.lydsy.com/JudgeOnline/problem.php?id=1051

    强连通分量,缩完点之后看出度为0的强连通分量有几个,如果只有一个则输出该强连通分量的点数,否则输出0;

    /**************************************************************
        Problem: 1051
        User: BLADEVIL
        Language: Pascal
        Result: Accepted
        Time:124 ms
        Memory:1396 kb
    ****************************************************************/
     
    //By BLADEVIL
    var
        n, m                        :longint;
        pre, other                  :array[0..100010] of longint;
        last                        :array[0..20010] of longint;
        l                           :longint;
        flag                        :array[0..10010] of boolean;
        dfn, low                    :array[0..10010] of longint;
        time                        :longint;
        tot                         :longint;
        stack                       :array[0..10010] of longint;
        key                         :array[0..10010] of longint;
        size                        :array[0..10010] of longint;
        full                        :longint;
        father                      :array[0..20010] of longint;
         
    function min(a,b:longint):longint;
    begin
        if a>b then min:=b else min:=a;
    end;   
     
    procedure connect(x,y:longint);
    begin
        inc(l);
        pre[l]:=last[x];
        last[x]:=l;
        other[l]:=y;
    end;
         
    procedure init;
    var
        i                           :longint;
        x, y                        :longint;
    begin
        read(n,m);
        for i:=1 to m do
        begin
            read(x,y);
            connect(y,x);
        end;
    end;
     
    procedure dfs(x:longint);
    var
        q, p                        :longint;
        cur                         :longint;
    begin
        inc(time);
        dfn[x]:=time;
        low[x]:=time;
        flag[x]:=true;
        inc(tot);
        stack[tot]:=x;
         
        q:=last[x];
        while q<>0 do
        begin
            p:=other[q];
            if dfn[p]=0 then
            begin
                dfs(p);
                low[x]:=min(low[x],low[p]);
            end else
            if flag[p] then low[x]:=min(low[x],dfn[p]);
            q:=pre[q];
        end;
         
        cur:=-1;
        if dfn[x]=low[x] then
        begin
            while cur<>x do
            begin
                cur:=stack[tot];
                dec(tot);
                flag[cur]:=false;
                key[cur]:=x+n;
                inc(size[key[cur]]);
            end;
        end;
    end;
     
    procedure main;
    var
        i                           :longint;
        q, p                        :longint;
    begin
        for i:=1 to n do if dfn[i]=0 then dfs(i);
        for i:=1 to n do
        begin
            q:=last[i];
            while q<>0 do
            begin
                p:=other[q];
                if key[i]<>key[p] then
                begin
                    connect(key[i],key[p]);
                    father[key[p]]:=key[i];
                end;
                q:=pre[q];
            end;
        end;
        full:=-1;
        for i:=1 to n do
        begin
            if (father[key[i]]=0) and (full=-1) then full:=key[i];
            if (father[key[i]]=0) and (full<>-1) and (full<>key[i]) then
            begin
                writeln(0);
                exit;
            end;
        end;
        writeln(size[full]);
    end;
     
    begin
        init;
        main;
    end.
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  • 原文地址:https://www.cnblogs.com/BLADEVIL/p/3433533.html
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