Bzoj 1069 [SCOI2007] maximum land area rotation jam

Test Instructions:Link Method:Rotating Jam parsing:It's like the POJ2079 triangle. So the idea is actually a train of thought. But notice that the subject is to look for the four-sided shape, if you find three moving points, it seems unrealistic. So we enumerate diagonal words, there must be a point diagonally up and down, and found the point below the counter-clockwise rotation, the above point is actually counterclockwise. So according to the idea of rotating jam, we need to enumerate the diagonal, and then find two dynamic point update maximum. This complexity is O (n^2)-enumerated fixed point and its diagonal point, for the subject can be too. Code:

#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define N 50010#define EPS 1e-8using namespace STD;intNstructpoint{Doublex, y;friendistream&operator>> (IStream &_,point &a) {scanf("%LF%LF", &AMP;A.X,&AMP;A.Y);return_; } point () {}, point (DoubleADoubleb): X (a), Y (b) {} pointoperator+ (ConstPoint &a) {returnPoint (X+A.X,Y+A.Y);} Pointoperator- (ConstPoint &a) {returnPoint (X-A.X,Y-A.Y);} Pointoperator* (DoubleRate) {returnPoint (X*rate,y*rate);}Double operator* (ConstPoint &a) {returnX*A.X+Y*A.Y;}Double operator^ (ConstPoint &a) {returnx*a.y-y*a.x;}friend BOOL operator< (point A,point b) {if(a.x==b.x)returna.y>b.y;returna.x<b.x; }}pt[n];intSta[n];intStaa[n];intTopintToppDoubleRotating_calipers () {Doubleret=0; for(intI=0; i<topp;i++) {intp=1, q=1; for(intj=i+2; j<topp;j++) { while(((pt[staa[p+1]]-pt[staa[i]] ^ (pt[staa[j]]-pt[staa[i])) > ((pt[staa[p]]-pt[staa[i]) ^ (pt[staa[j]]-pt[staa[i]))) P= (p+1)%topp; while(((pt[staa[j]]-pt[staa[i)] ^ (pt[staa[q+1]]-pt[staa[i]]) > (pt[staa[j]]-pt[staa[i]) ^ (pt[staa[q]]-pt[staa[i])) q= (q+1)%topp; Ret=max (ret, ((Pt[staa[p]]-pt[staa[i]) ^ (Pt[staa[j]]-pt[staa[i])) + ((Pt[staa[j]]-pt[staa[i]) ^ (pt[staa[q]]-pt[s        Taa[i])); }    }returnRET;}intMain () { while(~scanf("%d", &n) &&n!=-1)    { for(intI=1; i<=n;i++)Cin>>pt[i]; Sort (pt+1, pt+n+1); top=0, topp=-1; for(intI=1; i<=n;i++) { while(top>1&& ((Pt[i]-pt[sta[top]) ^ (pt[sta[top]]-pt[sta[top-1]]) >=0) top--;        Sta[++top]=i; } for(intI=1; i<top;i++) {staa[++topp]=sta[i]; } top=0; for(intI=1; i<=n;i++) { while(top>1&& ((Pt[i]-pt[sta[top]) ^ (pt[sta[top]]-pt[sta[top-1]]) <=0) top--;        Sta[++top]=i; } for(inti=top;i>=1; i--) {staa[++topp]=sta[i]; }printf("%.3lf\n", Rotating_calipers ()/2.0); }}

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Bzoj 1069 [SCOI2007] maximum land area rotation jam