【POJ2482】Stars in Your Window（线段树，扫描线）

题意：在二维坐标系中有一些带权值的点，要求用一个长宽指定不能互换的框套住其中的一些，使得它们的权值和最大。

n<=10000 x,y<=2^31

思路：首先按X排序，将Y坐标离散化，X坐标用扫描线框定，每个点(x,y)在x中只对y有a[i]的贡献，y+h有-a[i]的贡献，线段树（树状数组更好写）维护最大子段和即可。

  1 var t:array[1..200000]of record
  2                            l,r,s,m:int64;
  3                           end;
  4     x,y,c,a,h:array[1..50000]of int64;
  5     n,m,i,j,tt,ww,up,w1,h1:longint;
  6     ans:int64;
  7
  8 procedure swap(var x,y:int64);
  9 var t:int64;
 10 begin
 11  t:=x; x:=y; y:=t;
 12 end;
 13
 14 procedure qsort(l,r:longint);
 15 var i,j,mid1,mid2:longint;
 16 begin
 17  i:=l; j:=r; mid1:=x[(l+r)>>1]; mid2:=y[(l+r)>>1];
 18  repeat
 19   while (mid1>x[i])or((mid1=x[i])and(mid2>y[i])) do inc(i);
 20   while (mid1<x[j])or((mid1=x[j])and(mid2<y[j])) do dec(j);
 21   if i<=j then
 22   begin
 23    swap(x[i],x[j]);
 24    swap(y[i],y[j]);
 25    swap(a[i],a[j]);
 26    inc(i); dec(j);
 27   end;
 28  until i>j;
 29  if l<j then qsort(l,j);
 30  if i<r then qsort(i,r);
 31 end;
 32
 33 procedure qsort2(l,r:longint);
 34 var i,j:longint;
 35     mid:int64;
 36 begin
 37  i:=l; j:=r; mid:=c[(l+r)>>1];
 38  repeat
 39   while mid>c[i] do inc(i);
 40   while mid<c[j] do dec(j);
 41   if i<=j then
 42   begin
 43    swap(c[i],c[j]);
 44    inc(i); dec(j);
 45   end;
 46  until i>j;
 47  if l<j then qsort2(l,j);
 48  if i<r then qsort2(i,r);
 49 end;
 50
 51 function max(x,y:int64):int64;
 52 begin
 53  if x>y then exit(x);
 54  exit(y);
 55 end;
 56
 57 procedure pushup(p:longint);
 58 var ls,rs:longint;
 59 begin
 60  ls:=p<<1; rs:=ls+1;
 61  t[p].l:=max(t[ls].l,t[ls].s+t[rs].l);
 62  t[p].r:=max(t[rs].r,t[rs].s+t[ls].r);
 63  t[p].m:=max(t[ls].r+t[rs].l,max(t[ls].m,t[rs].m));
 64 end;
 65
 66 procedure update(l,r,x,v,p:longint);
 67 var mid:longint;
 68 begin
 69  if l=r then
 70  begin
 71   t[p].s:=t[p].s+v;
 72   t[p].l:=t[p].s; t[p].r:=t[p].s; t[p].m:=t[p].s;
 73   exit;
 74  end;
 75  mid:=(l+r)>>1;
 76  if x<=mid then update(l,mid,x,v,p<<1)
 77   else update(mid+1,r,x,v,p<<1+1);
 78  t[p].s:=t[p].s+v;
 79  pushup(p);
 80 end;
 81
 82 function hash(x:int64):longint;
 83 var l,r,mid,last:longint;
 84 begin
 85  l:=1; r:=up; last:=1;
 86  while l<=r do
 87  begin
 88   mid:=(l+r)>>1;
 89   if x=h[mid] then begin last:=mid; r:=mid-1; end;
 90   if x<h[mid] then r:=mid-1;
 91   if x>h[mid] then l:=mid+1;
 92  end;
 93  exit(last);
 94 end;
 95
 96 begin
 97  assign(input,‘poj2482.in‘); reset(input);
 98  assign(output,‘poj2482.out‘); rewrite(output);
 99  while not eof do
100  begin
101   readln(n,w1,h1);
102   if n=0 then break;
103   for i:=1 to n do read(x[i],y[i],a[i]);
104   qsort(1,n);
105   m:=0;
106   for i:=1 to n do
107   begin
108    inc(m); c[m]:=y[i];
109    inc(m); c[m]:=y[i]+h1;
110   end;
111   qsort2(1,m);
112   up:=1; h[1]:=c[1];
113   for i:=2 to m do
114    if c[i]>c[i-1] then
115    begin
116     if c[i]=c[i-1]+1 then begin inc(up); h[up]:=c[i]; end
117      else
118      begin
119       inc(up); h[up]:=c[i-1];
120       inc(up); h[up]:=c[i];
121      end;
122    end;
123   ans:=-maxlongint;
124   tt:=1; ww:=0;
125   while ww<n do
126   begin
127    inc(ww);
128    while (tt<=n)and(x[tt]+w1-1<x[ww]) do
129    begin
130     update(1,up,hash(y[tt]),-a[tt],1);
131     update(1,up,hash(y[tt]+h1),a[tt],1);
132     inc(tt);
133    end;
134    update(1,up,hash(y[ww]),a[ww],1);
135    update(1,up,hash(y[ww]+h1),-a[ww],1);
136    ans:=max(ans,t[1].m);
137   end;
138   writeln(ans);
139   for i:=1 to up<<2 do
140   begin
141    t[i].s:=0; t[i].l:=0; t[i].r:=0; t[i].m:=0;
142   end;
143  end;
144  close(input);
145  close(output);
146 end.