BZOJ1069: [SCOI2007]最大土地面积
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对于给定的点,先求出凸包,听说水平序求凸包会被卡..亲测不会
然后对于求出来的凸包,求出每一个对踵点。然后对于每一个对踵点,遍历凸包上每一个点,求出最大的叉积和最小的叉积,绝对值的累加即位最大面积。
//BZOJ1069 //by Cydiater //2017.1.29 #include <iostream> #include <queue> #include <map> #include <cstring> #include <string> #include <cstdlib> #include <cmath> #include <ctime> #include <iomanip> #include <algorithm> #include <cstdio> #include <bitset> #include <set> #include <vector> using namespace std; #define ll long long #define up(i,j,n) for(int i=j;i<=n;i++) #define down(i,j,n) for(int i=j;i>=n;i--) #define cmax(a,b) a=max(a,b) #define cmin(a,b) a=min(a,b) #define db double #define Vector Point const int MAXN=1e4+5; const int oo=0x3f3f3f3f; const db eps=1e-10; struct Point{ db x,y; Point(db x=0,db y=0):x(x),y(y){} }; Vector operator + (Point x,Point y){return Vector(x.x+y.x,x.y+y.y);} Vector operator - (Point x,Point y){return Vector(x.x-y.x,x.y-y.y);} Vector operator * (Vector x,db p){return Vector(x.x*p,x.y*p);} Vector operator / (Vector x,db p){return Vector(x.x/p,x.y/p);} int dcmp(db x){if(fabs(x)<eps)return 0;else return x<0?-1:1;} bool operator < (const Vector &x,const Vector &y){return dcmp(x.x-y.x)==0?x.y<y.y:x.x<y.x;} bool operator == (const Vector &x,const Vector &y){return dcmp(x.x-y.x)==0&&dcmp(x.y-y.y)==0;} int N,top=0; Point V[MAXN],q[MAXN]; db ans=0; namespace solution{ Point Write(){db x,y;scanf("%lf%lf",&x,&y);return Point(x,y);} db Cross(Vector x,Vector y){return x.x*y.y-x.y*y.x;} void Prepare(){ scanf("%d",&N); up(i,1,N)V[i]=Write(); sort(V+1,V+N+1); N=unique(V+1,V+N+1)-(V+1); } void Andrew(){ up(i,1,N){ while(top>=2&&dcmp(Cross(q[top]-q[top-1],V[i]-q[top-1]))<=0)top--; q[++top]=V[i]; } int lim=top; down(i,N-1,1){ while(top-lim>=1&&dcmp(Cross(q[top]-q[top-1],V[i]-q[top-1]))<=0)top--; q[++top]=V[i]; } } db Area(Point A,Point B){ db area1=0,area2=0; up(i,1,top){ cmax(area1,Cross(A-q[i],B-q[i])/2); cmin(area2,Cross(A-q[i],B-q[i])/2); } return area1-area2; } void Solve(){ Andrew(); top--; int pos=2; up(i,1,top){ while(fabs(Cross(q[pos+1]-q[i+1],q[i]-q[i+1]))>fabs(Cross(q[pos]-q[i+1],q[i]-q[i+1]))){ pos%=top;pos++; } cmax(ans,Area(q[pos],q[i])); } printf("%.3lf\n",ans); } } int main(){ //freopen("input.in","r",stdin); using namespace solution; Prepare(); Solve(); return 0; }
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