[几何 two-pointers] BZOJ 1278 向量vector

来源：互联网发布：穿淘宝爆款编辑：程序博客网时间：2024/09/21 08:52

有一个结论就是组成答案的向量一定都在某一直线的一侧
证明：以答案向量作一个圆这条直线就是过终点的切线

#include<cstdio>#include<cstdlib>#include<algorithm>#include<cmath>using namespace std;typedef long long ld;struct PP{  ld x,y; double ang;  PP(ld x=0,ld y=0):x(x),y(y) { }  friend PP operator + (PP A,PP B){ return PP(A.x+B.x,A.y+B.y); }  friend PP operator - (PP A,PP B){ return PP(A.x-B.x,A.y-B.y); }  friend ld operator * (PP A,PP B){ return A.x*B.y-B.x*A.y; }  void Ang() { ang=atan2(y,x); }  bool operator < (const PP &B) const{    return ang<B.ang;  }  ld M(){ return x*x+y*y; }};const int N=200005;int n;PP p[N],sum[N];ld Ans=0;int main(){  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  scanf("%d",&n);  for (int i=1;i<=n;i++) scanf("%lld%lld",&p[i].x,&p[i].y),p[i].Ang();  sort(p+1,p+n+1);  for (int i=1;i<=n;i++) p[n+i]=p[i];  for (int i=1;i<=(n<<1);i++) sum[i]=sum[i-1]+p[i];  int l=1,r=1;  Ans=max(Ans,(sum[r]-sum[l-1]).M());  while (r+1<=l+n-1 && p[l]*p[r+1]>=0) r++,Ans=max(Ans,(sum[r]-sum[l-1]).M());  Ans=max(Ans,(sum[r]-sum[l-1]).M());  while (l+1<=n){    l++;     Ans=max(Ans,(sum[r]-sum[l-1]).M());    while (r+1<=l+n-1 && p[l]*p[r+1]>=0) r++,Ans=max(Ans,(sum[r]-sum[l-1]).M());    Ans=max(Ans,(sum[r]-sum[l-1]).M());  }  printf("%lld.000\n",Ans);  return 0;}

0 0