[BZOJ1137][POI2009]Wsp 岛屿（半平面交）

2019-11-08 03:27:45

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题目描述

传送门

题解

这道题路径的交点处是可以随意通行的如果1->n是可通行的那么直接走就行了如果不能通行对于每一个点i，找出它能直接走到的编号最大的点，显然只有这个点是有用的然后从点i向这个点连一条直线，加上n->1这条直线，实际上交出了一个凸多边形答案即为这个凸多边形的边长-1->n的路径长

代码

#include<algorithm>#include<iostream>#include<cstring>#include<cstdio>#include<cmath>using namespace std;#define N 1000005const double eps=1e-9;const double inf=1e7;int dcmp(double x){ if (x<=eps&&x>=-eps) return 0; return (x>0)?1:-1;}struct data{ int x,y; bool Operator < (const data &a) const { return x<a.x||x==a.x&&y>a.y; }}e[N];struct Vector{ double x,y; Vector(double X=0,double Y=0,double ANG=0) { x=X,y=Y; }};typedef Vector Point;struct Line{ Point p; Vector v; double ang; Line(Point P=Point(0,0),Vector V=Vector(0,0),double ANG=0) { p=P,v=V,ang=atan2(v.y,v.x); } bool operator < (const Line &a) const { return ang<a.ang; }};Vector operator + (Vector a,Vector b) {return Vector(a.x+b.x,a.y+b.y);}Vector operator - (Vector a,Vector b) {return Vector(a.x-b.x,a.y-b.y);}Vector operator * (Vector a,double b) {return Vector(a.x*b,a.y*b);}int n,m,cnt,l,r,est[N];double ans;Point pt[N],p[N],poly[N];Line L[N],q[N];bool flag;double Len(Vector a){ return sqrt(a.x*a.x+a.y*a.y);}double Cross(Vector a,Vector b){ return a.x*b.y-a.y*b.x;}Point GLI(Point P,Vector v,Point Q,Vector w){ Vector u=P-Q; double t=Cross(w,u)/Cross(v,w); return P+v*t;}bool Onleft(Line m,Point P){ Vector w=P-m.p; return dcmp(Cross(m.v,w))>=0;}void halfp(){ sort(L+1,L+cnt+1); q[l=r=1]=L[1]; for (int i=2;i<=cnt;++i) { while (l<r&&!Onleft(L[i],p[r-1])) --r; while (l<r&&!Onleft(L[i],p[l])) ++l; q[++r]=L[i]; if (dcmp(Cross(q[r].v,q[r-1].v))==0) { --r; if (Onleft(q[r],L[i].p)) q[r]=L[i]; } if (l<r) p[r-1]=GLI(q[r-1].p,q[r-1].v,q[r].p,q[r].v); } while (l<r&&!Onleft(q[l],p[r-1])) --r; cnt=0; if (r-l<=1) return; p[r]=GLI(q[r].p,q[r].v,q[l].p,q[l].v); for (int i=l;i<=r;++i) poly[++cnt]=p[i];}int main(){ scanf("%d%d",&n,&m); for (int i=1;i<=n;++i) scanf("%lf%lf",&pt[i].x,&pt[i].y); for (int i=1;i<=m;++i) { scanf("%d%d",&e[i].x,&e[i].y); if (e[i].x>e[i].y) swap(e[i].x,e[i].y); if (e[i].x==1&&e[i].y==n) flag=1; } if (!flag) { ans=Len(pt[1]-pt[n]); PRintf("%.9lf/n",ans); } sort(e+1,e+m+1); for (int i=1;i<=n;++i) est[i]=n+1; int now=1; while (now<=m) { int i; for (i=now;e[i].x==e[now].x;++i) { if (i==now&&e[i].y!=n) {est[e[now].x]=n;break;} else if (i!=now&&e[i].y+1!=e[i-1].y) {est[e[now].x]=e[i-1].y-1;break;} } for (;e[i].x==e[now].x;++i); if (est[e[now].x]==n+1) { est[e[now].x]=e[i-1].y-1; if (e[now].x==e[i-1].y-1) --est[e[now].x]; } now=i; } L[++cnt]=Line(Point(inf,inf),Vector(-2*inf,0)); L[++cnt]=Line(Point(-inf,inf),Vector(0,-2*inf)); L[++cnt]=Line(Point(-inf,-inf),Vector(2*inf,0)); L[++cnt]=Line(Point(inf,-inf),Vector(0,2*inf)); L[++cnt]=Line(pt[1],pt[n]-pt[1]); for (int i=1;i<n;++i) { if (est[i]==n+1) est[i]=n; if (est[i]&&est[i]>i) L[++cnt]=Line(pt[est[i]],pt[i]-pt[est[i]]); } halfp(); for (int i=1;i<=cnt;++i) ans+=Len(poly[i%cnt+1]-poly[(i+1)%cnt+1]); ans-=Len(pt[1]-pt[n]); printf("%.9lf/n",ans);}

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