题意:
给平面上n个点,求这n个点组成的最大三角形面积。
分析:
旋转卡壳,但要注意和求平面最远点对的区别,最大三角形的边不一定在凸包上的,也贴出以前写的求平面最远点对的poj 2187代码作为对比。
代码:
//poj 2079 //sep9 #include <iostream> #include <algorithm> using namespace std; const int maxN=50012; struct P { int x,y; }pnt[maxN],cnt[maxN]; int n; int cmp(P a,P b) { if(a.y!=b.y) return a.y<b.y; return a.x<b.x; } int cross(P a,P b,P c) { int x1=b.x-a.x; int y1=b.y-a.y; int x2=c.x-a.x; int y2=c.y-a.y; return x1*y2-x2*y1; } int graham() { sort(pnt,pnt+n,cmp); int i,pos=1; cnt[0]=pnt[0]; cnt[1]=pnt[1]; for(i=2;i<n;++i){ while(pos>0&&cross(cnt[pos-1],cnt[pos],pnt[i])<=0) --pos; cnt[++pos]=pnt[i]; } int bak=pos; for(i=n-2;i>=0;--i){ while(pos>bak&&cross(cnt[pos-1],cnt[pos],pnt[i])<=0) --pos; cnt[++pos]=pnt[i]; } return pos; } int main() { while(scanf("%d",&n)==1&&n!=-1){ for(int i=0;i<n;++i) scanf("%d%d",&pnt[i].x,&pnt[i].y); int pos=graham(); int ans=0,p=1,i,j,k; for(i=j=k=0;i<pos;++i){ while(abs(cross(cnt[i],cnt[j],cnt[(k+1)%pos]))>abs(cross(cnt[i],cnt[j],cnt[k]))) k=(k+1)%pos; while(abs(cross(cnt[i],cnt[(j+1)%pos],cnt[k]))>abs(cross(cnt[i],cnt[j],cnt[k]))) j=(j+1)%pos; ans=max(ans,abs(cross(cnt[i],cnt[j],cnt[k]))); } printf("%.2lf\n",ans/2.0); } return 0; }
求最远点对的代码:
//poj 2187 //sep9 #include<iostream> #include<algorithm> using namespace std; struct Point { int x,y; }points[50012],bag[50012]; int cross(Point p1,Point p2,Point p0) { return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y); } int cmp(Point a,Point b) { return a.y==b.y?a.x<b.x:a.y<b.y; } int dis(Point a,Point b) { return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y); } int getBag(Point p[],int n,Point bag[]) { int i,bp=1; bag[0]=p[0]; bag[1]=p[1]; for(i=2;i<n;++i) { while(bp&&cross(bag[bp-1],p[i],bag[bp])<=0) --bp; bag[++bp]=p[i]; } int len=bp; bag[++bp]=p[n-2]; for(i=n-3;i>=0;--i) { while(bp>len&&cross(bag[bp-1],p[i],bag[bp])<=0) --bp; bag[++bp]=p[i]; } return bp; } int main() { int p,i,n; while(scanf("%d",&n)==1) { for(i=0;i<n;++i) scanf("%d%d",&points[i].x,&points[i].y); sort(points,points+n,cmp); int dif,ans=0,m=getBag(points,n,bag); p=1; for(i=0;i<m;++i) { while(abs(cross(bag[i],bag[i+1],bag[p+1]))>abs(cross(bag[i],bag[i+1],bag[p]))) p=(p+1)%m; ans=max(ans,max(dis(bag[i],bag[p]),dis(bag[i+1],bag[p+1]))); } printf("%d\n",ans); } }
时间: 2024-10-10 12:01:18