有m个国家(m<=20)对每个国家给定n个城镇 这个国家的围墙是保证围住n个城镇的周长最短的多边形 必然是凸包
进行若干次导弹发射 落到一个国家内则国家被破坏
最后回答总共有多少面积被破坏
首先求凸包
然后判断点是否在凸包内 要用O(logn)的判断方法 不然会超时
这道题常数卡的有点紧 TLE三次才过
蒟蒻没救了 2017年做1991年的题还被卡常数
#include<iostream> #include<stdio.h> #include<stdlib.h> #include<string.h> #include<math.h> #include<algorithm> #include<queue> #include<vector> using namespace std; const double eps=1e-9; int cmp(double x) { if(fabs(x)<eps)return 0; if(x>0)return 1; else return -1; } const double pi=acos(-1.0); inline double sqr(double x) { return x*x; } struct point { double x,y; point (){} point (double a,double b):x(a),y(b){} void input() { scanf("%lf%lf",&x,&y); } friend point operator +(const point &a,const point &b) { return point(a.x+b.x,a.y+b.y); } friend point operator -(const point &a,const point &b) { return point(a.x-b.x,a.y-b.y); } friend bool operator ==(const point &a,const point &b) { return cmp(a.x-b.x)==0&&cmp(a.y-b.y)==0; } friend point operator *(const point &a,const double &b) { return point(a.x*b,a.y*b); } friend point operator*(const double &a,const point &b) { return point(a*b.x,a*b.y); } friend point operator /(const point &a,const double &b) { return point(a.x/b,a.y/b); } double norm() { return sqrt(sqr(x)+sqr(y)); } }; struct line { point a,b; line(){}; line(point x,point y):a(x),b(y) { } }; double det(const point &a,const point &b) { return a.x*b.y-a.y*b.x; } double dot(const point &a,const point &b) { return a.x*b.x+a.y*b.y; } double dist(const point &a,const point &b) { return (a-b).norm(); } point rotate_point(const point &p,double A) { double tx=p.x,ty=p.y; return point(tx*cos(A)-ty*sin(A),tx*sin(A)+ty*cos(A)); } bool parallel(line a,line b) { return !cmp(det(a.a-a.b,b.a-b.b)); } bool line_joined(line a,line b,point &res) { if(parallel(a,b))return false; double s1=det(a.a-b.a,b.b-b.a); double s2=det(a.b-b.a,b.b-b.a); res=(s1*a.b-s2*a.a)/(s1-s2); return true; } bool pointonSegment(point p,point s,point t) { return cmp(det(p-s,t-s))==0&&cmp(dot(p-s,p-t))<=0; } void PointProjLine(const point p,const point s,const point t,point &cp) { double r=dot((t-s),(p-s))/dot(t-s,t-s); cp=s+r*(t-s); } struct polygon_convex { vector<point>P; polygon_convex(int Size=0) { P.resize(Size); } }; bool comp_less(const point &a,const point &b) { return cmp(a.x-b.x)<0||cmp(a.x-b.x)==0&&cmp(a.y-b.y)<0; } polygon_convex convex_hull(vector<point> a) { polygon_convex res(2*a.size()+5); sort(a.begin(),a.end(),comp_less); a.erase(unique(a.begin(),a.end()),a.end());//删去重复点 int m=0; for(int i=0;i<a.size();i++) { while(m>1&&cmp(det(res.P[m-1]-res.P[m-2],a[i]-res.P[m-2]))<=0)--m; res.P[m++]=a[i]; } int k=m; for(int i=int(a.size())-2;i>=0;--i) { while(m>k&&cmp(det(res.P[m-1]-res.P[m-2],a[i]-res.P[m-2]))<=0)--m; res.P[m++]=a[i]; } res.P.resize(m); if(a.size()>1)res.P.resize(m-1); return res; } bool is_convex(vector<point> &a) { for(int i=0;i<a.size();i++) { int i1=(i+1)%int(a.size()); int i2=(i+2)%int(a.size()); int i3=(i+3)%int(a.size()); if((cmp(det(a[i1]-a[i],a[i2]-a[i1]))*cmp(det(a[i2]-a[i1],a[i3]-a[i2])))<0) return false; } return true; } int containO(const polygon_convex &a,const point &b) { int n=a.P.size(); point g=(a.P[0]+a.P[n/3]+a.P[2*n/3])/3.0; int l=0,r=n; while(l+1<r) { int mid=(l+r)/2; if(cmp(det(a.P[l]-g,a.P[mid]-g))>0) { if(cmp(det(a.P[l]-g,b-g))>=0&&cmp(det(a.P[mid]-g,b-g))<0)r=mid; else l=mid; }else { if(cmp(det(a.P[l]-g,b-g))<0&&cmp(det(a.P[mid]-g,b-g))>=0)l=mid; else r=mid; } } r%=n; int z=cmp(det(a.P[r]-b,a.P[l]-b))-1; if(z==-2)return 1; return z; } polygon_convex pc[30]; double area(int n) { double ans=0; vector<point>a; a.clear(); for(int i=0;i<pc[n].P.size();i++) a.push_back(pc[n].P[i]); a.push_back(a[0]); for(int i=0;i<a.size()-1;i++)ans+=det(a[i+1],a[i]); //cout<<ans/2<<endl; return ans/2.0; } int tot; double are[30]; vector<point> pp; int shoot[600][600]; bool damed[30]; int main() {//freopen("t.txt","r",stdin); int n; tot=1; while(scanf("%d",&n)&&n!=-1) { pp.clear(); for(int i=0;i<n;i++) { point p; p.input(); pp.push_back(p); } pc[tot]=convex_hull(pp); //cout<<pc[tot].P.size()<<endl; are[tot]=area(tot); //cout<<are[tot]<<endl; tot++; } int x,y; memset(damed,0,sizeof(damed)); int sum=0; memset(shoot,0,sizeof(shoot)); while(scanf("%d%d",&x,&y)!=EOF) { if(sum==tot-1)break; if(shoot[x][y]==0) { for(int i=1;i<tot;i++) if(containO(pc[i],point(x,y))) { shoot[x][y]=i; break; } } if(!damed[shoot[x][y]]){sum++;damed[shoot[x][y]]=1;} } double tarea=0; for(int i=1;i<tot;i++) { //cout<<are[i]<<endl; if(damed[i])tarea+=are[i]; } printf("%.2lf\n",-tarea+0.0005); return 0; }
时间: 2024-10-20 10:08:22