【BZOJ3295】动态逆序对（BIT套动态加点线段树）

题意：对于序列A，它的逆序对数定义为满足i<j，且Ai>Aj的数对(i,j)的个数。

给1到n的一个排列，按照某种顺序依次删除m个元素，你的任务是在每次删除一个元素之前统计整个序列的逆序对数。

N<=100000 M<=50000

思路：树套树即可，第i棵树维护前i位中1-j的个数，支持区间求和

外面要套BIT来实现区间加

本来维护剩下的数就可以，但初始化可能要开(n log^2 n) 个个节点，这题又只有128MB……

所以只能维护删除的数了，如果256MB乱搞就行了……

 

  1 var t:array[0..5100000,0..1]of longint;
  2     sum:array[0..5100000]of int64;
  3     root,a,b,a1,a2,c:array[0..110000]of longint;
  4     n,m,i,x,cnt,l,r,j:longint;
  5     ans:int64;
  6 
  7 
  8 procedure pushup(x:longint);
  9 begin
 10  l:=t[x,0]; r:=t[x,1];
 11  sum[x]:=sum[l]+sum[r];
 12 end;
 13 
 14 procedure update(l,r,x,v:longint;var p:longint);
 15 var mid:longint;
 16 begin
 17  if p=0 then
 18  begin
 19   inc(cnt); p:=cnt;
 20  end;
 21  if l=r then
 22  begin
 23   sum[p]:=sum[p]+v; exit;
 24  end;
 25  mid:=(l+r)>>1;
 26  if x<=mid then update(l,mid,x,v,t[p,0]);
 27  if x>mid then update(mid+1,r,x,v,t[p,1]);
 28  pushup(p);
 29 end;
 30 
 31 function query(l,r,x,y,p:longint):int64;
 32 var mid:longint;
 33 begin
 34  if p=0 then exit(0);
 35  if (l>=x)and(r<=y) then exit(sum[p]);
 36  mid:=(l+r)>>1;
 37  query:=0;
 38  if x<=mid then query:=query+query(l,mid,x,y,t[p,0]);
 39  if y>mid then query:=query+query(mid+1,r,x,y,t[p,1]);
 40  pushup(p);
 41 end;
 42 
 43 function lowbit(x:longint):longint;
 44 begin
 45  exit(x and (-x));
 46 end;
 47 
 48 procedure change(i,j,v:longint);
 49 var k:longint;
 50 begin
 51  k:=i;
 52  while k<=n do
 53  begin
 54   update(1,n,j,v,root[k]);
 55   k:=k+lowbit(k);
 56  end;
 57 end;
 58 
 59 function getsum(x:longint):longint;
 60 begin
 61  getsum:=0;
 62  while x>0 do
 63  begin
 64   getsum:=getsum+c[x];
 65   x:=x-lowbit(x);
 66  end;
 67 end;
 68 
 69 function ask(i,j,x,y:longint):int64;
 70 var k:longint;
 71 begin
 72  ask:=0;
 73  if i>j then exit(0);
 74  if x>y then exit(0);
 75  k:=j;
 76  while k>0 do
 77  begin
 78   ask:=ask+query(1,n,x,y,root[k]);
 79   k:=k-lowbit(k);
 80  end;
 81  k:=i-1;
 82  while k>0 do
 83  begin
 84   ask:=ask-query(1,n,x,y,root[k]);
 85   k:=k-lowbit(k);
 86  end;
 87 end;
 88 
 89 begin
 90  assign(input,'bzoj3295.in'); reset(input);
 91  assign(output,'bzoj3295.out'); rewrite(output);
 92  readln(n,m);
 93 
 94  for i:=1 to n do
 95  begin
 96   read(a[i]);
 97   b[a[i]]:=i;
 98   a1[i]:=getsum(n)-getsum(a[i]);
 99   ans:=ans+a1[i];
100   j:=a[i];
101   while j<=n do
102   begin
103    inc(c[j]);
104    j:=j+lowbit(j);
105   end;
106  end;
107 
108  fillchar(c,sizeof(c),0);
109  for i:=n downto 1 do
110  begin
111   a2[i]:=getsum(a[i]-1);
112   j:=a[i];
113   while j<=n do
114   begin
115    inc(c[j]);
116    j:=j+lowbit(j);
117   end;
118  end;
119 
120 { for i:=1 to n do write(a1[i],' ');
121  writeln;
122  for i:=1 to n do write(a2[i],' ');  }
123  for i:=1 to m do
124  begin
125   readln(x);
126   writeln(ans);
127   j:=b[x];
128   ans:=ans-(a1[j]+a2[j]-ask(1,j-1,x+1,n)-ask(j+1,n,1,x-1));
129   change(j,x,1);
130  end;
131 
132 
133  close(input);
134  close(output);
135 end.

 

posted on 2017-02-23 19:20  myx12345  阅读(196)  评论(0)  编辑  收藏  举报

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