bzoj 1007 凸壳
首先明确一个概念
左面内个叫上凸壳,右面那个叫下凸壳
然后我们只需要维护一个上图壳就行了,先按着斜率排序,每次加进来一条边,判断tot边和这个边与tot-1边的交点横坐标,
如果这条边的横坐标小就一直弹栈就好了
/************************************************************** Problem: 1007 User: BLADEVIL Language: Pascal Result: Accepted Time:360 ms Memory:3352 kb ****************************************************************/ //By BLADEVIL var n :longint; a, b, num :array[0..100010] of longint; quea, queb :array[0..100010] of longint; tot :longint; quex :array[0..100010] of double; ans :array[0..100010] of longint; i :longint; procedure swap(var a,b:longint); var c :longint; begin c:=a; a:=b; b:=c; end; procedure qs(low,high:longint); var i, j, xx, yy :longint; begin i:=low; j:=high; xx:=a[(i+j) div 2]; yy:=b[(i+j) div 2]; while i<j do begin while (a[i]<xx) or (a[i]=xx) and (b[i]>yy) do inc(i); while (a[j]>xx) or (a[j]=xx) and (b[j]<yy) do dec(j); if i<=j then begin swap(a[i],a[j]); swap(b[i],b[j]); swap(num[i],num[j]); inc(i); dec(j); end; end; if i<high then qs(i,high); if j>low then qs(low,j); end; procedure qs1(low,high:longint); var i, j, xx :longint; begin i:=low; j:=high; xx:=ans[(i+j) div 2]; while i<j do begin while ans[i]<xx do inc(i); while ans[j]>xx do dec(j); if i<=j then begin swap(ans[i],ans[j]); inc(i); dec(j); end; end; if i<high then qs1(i,high); if j>low then qs1(low,j); end; procedure insert(i:longint); var k :longint; x :double; begin if a[i]=quea[tot] then exit; if tot>1 then begin x:=(queb[tot-1]-b[i])/(a[i]-quea[tot-1]); while (tot>1) and (x<=quex[tot]) do begin dec(tot); x:=(queb[tot-1]-b[i])/(a[i]-quea[tot-1]); end; end; inc(tot); quea[tot]:=a[i]; queb[tot]:=b[i]; quex[tot]:=(queb[tot-1]-b[i])/(a[i]-quea[tot-1]); ans[tot]:=num[i]; end; begin read(n); for i:=1 to n do read(a[i],b[i]); for i:=1 to n do num[i]:=i; qs(1,n); quea[1]:=a[1]; queb[1]:=b[1]; quex[1]:=-maxlongint; ans[1]:=num[1]; tot:=1; for i:=2 to n do insert(i); qs1(1,tot); for i:=1 to tot do write(ans[i],' '); end.