【BZOJ2049】洞穴勘测（LCT）

题意：一张图，要求支持以下操作：

1.加边

2.删边

3.询问两点之间是否联通

100%的数据满足n≤10000, m≤200000 

思路：LCT裸题，不需要维护任何信息

  1 var t:array[0..500000,0..1]of longint;
  2     fa,rev,q:array[0..500000]of longint;
  3     n,m,i,x,y,k,j,top,s:longint;
  4     ch:string;
  5 
  6 procedure swap(var x,y:longint);
  7 var t:longint;
  8 begin
  9  t:=x; x:=y; y:=t;
 10 end;
 11 
 12 function isroot(x:longint):boolean;
 13 begin
 14  if (t[fa[x],0]<>x)and(t[fa[x],1]<>x) then exit(true);
 15  exit(false);
 16 end;
 17 
 18 procedure pushdown(x:longint);
 19 var l,r:longint;
 20 begin
 21  l:=t[x,0]; r:=t[x,1];
 22  if rev[x]>0 then
 23  begin
 24   rev[x]:=rev[x] xor 1; rev[l]:=rev[l] xor 1; rev[r]:=rev[r] xor 1;
 25   swap(t[x,0],t[x,1]);
 26  end;
 27 end;
 28 
 29 procedure rotate(x:longint);
 30 var y,z,l,r:longint;
 31 begin
 32  y:=fa[x]; z:=fa[y];
 33  if t[y,0]=x then l:=0
 34   else l:=1;
 35  r:=l xor 1;
 36  while not isroot(y) do
 37  begin
 38   if t[z,0]=y then t[z,0]:=x
 39    else t[z,1]:=x;
 40  end;
 41  fa[x]:=z; fa[y]:=x; fa[t[x,r]]:=y;
 42  t[y,l]:=t[x,r]; t[x,r]:=y;
 43 end;
 44 
 45 procedure splay(x:longint);
 46 var y,z,k:longint;
 47 begin
 48  inc(top); q[top]:=x;
 49  k:=x;
 50  while not isroot(k) do
 51  begin
 52   inc(top); q[top]:=fa[k];
 53   k:=fa[k];
 54  end;
 55 
 56  while top>0 do
 57  begin
 58   pushdown(q[top]);
 59   dec(top);
 60  end;
 61 
 62  while not isroot(x) do
 63  begin
 64   y:=fa[x]; z:=fa[y];
 65   if not isroot(y) then
 66   begin
 67    if (t[y,0]=x)xor(t[z,0]=y) then rotate(x)
 68     else rotate(y);
 69   end;
 70   rotate(x);
 71  end;
 72 end;
 73 
 74 procedure access(x:longint);
 75 var last:longint;
 76 begin
 77  last:=0;
 78  while x>0 do
 79  begin
 80   splay(x); t[x,1]:=last;
 81   last:=x; x:=fa[x];
 82  end;
 83 end;
 84 
 85 procedure makeroot(x:longint);
 86 begin
 87  access(x); splay(x); rev[x]:=rev[x] xor 1;
 88 end;
 89 
 90 procedure link(x,y:longint);
 91 begin
 92  makeroot(x); fa[x]:=y;
 93 end;
 94 
 95 procedure cut(x,y:longint);
 96 begin
 97  makeroot(x); access(y); splay(y); t[y,0]:=0; fa[x]:=0;
 98 end;
 99 
100 function findroot(x:longint):longint;
101 var k:longint;
102 begin
103  access(x); splay(x);
104  k:=x;
105  while t[k,0]<>0 do k:=t[k,0];
106  exit(k);
107 end;
108 
109 begin
110  assign(input,'bzoj2049.in'); reset(input);
111  assign(output,'bzoj2049.out'); rewrite(output);
112  readln(n,m);
113  for i:=1 to m do
114  begin
115   readln(ch); k:=length(ch); s:=0; x:=0; y:=0; j:=1;
116   repeat
117    if (ch[j]=' ') then
118    begin
119     inc(s);
120     while ch[j]=' ' do inc(j);
121     continue;
122    end;
123    if (ch[j]>='0')and(ch[j]<='9') then
124    begin
125     case s of
126      1:x:=x*10+ord(ch[j])-ord('0');
127      2:y:=y*10+ord(ch[j])-ord('0');
128     end;
129    end;
130    inc(j);
131   until j>k;
132   case ch[1] of
133    'C':link(x,y);
134    'D':cut(x,y);
135    'Q':
136     if findroot(x)<>findroot(y) then writeln('No')
137      else writeln('Yes');
138   end;
139  end;
140 
141 
142  close(input);
143  close(output);
144 end.

 

posted on 2017-02-12 20:22  myx12345  阅读(186)  评论(0)  编辑  收藏  举报

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