HDU 4766 Network(計算幾何 二分+三分)
阿新 • • 發佈:2019-02-06
這個題目標準解法估計不是二分+三分,不過我一見到這個題目就認為是二分或者三分,所以一直想下去就寫成了下面的程式碼,dubug好久才AC
二分列舉到房子的距離也就是路由器到房子的距離,然後三分判斷在以房子為圓心這個距離為半徑的的圓周上是否存在點能使這點到所有其他點
距離<r,這個三分判斷利用圓的凸性很好判斷,關鍵問題就是二分列舉距離的時候上下界無法確定,因為不具有單調性,但是如果能找到一點滿足
題目條件,以這點為二分列舉的右端點,那麼單調性就產生了。那麼怎麼找這個點呢,想起點集最小圓覆蓋,求出圓心與半徑(最小 是否存在問題
也解決了),最後二分!
#include <string.h> #include <stdio.h> #include <algorithm> #include <iostream> #include <math.h> using namespace std; #define MIN(a,b) (a<b?a:b) #define MAX(a,b) (a>b?a:b) #define inf 1e12 #define eps 1e-8 #define maxn 1500 const double PI=acos(-1.0); struct point{ double x,y; point(double _x=0,double _y=0):x(_x),y(_y){} }po[maxn],cir,temp,na; int n; double r,LL; double dis(point &a,point &b){ return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); } bool is_ok(){ double re=r*2,t=0,p; point pp; for(int i=1;i<=n;i++) for(int j=i+1;j<=n;j++){ p=dis(po[i],po[j]); if(p>t){ t=p; pp=point((po[i].x+po[j].x)/2,(po[i].y+po[j].y)/2); LL=dis(pp,cir); } } if(t>re) return false; return true; } int get_max_point(point &a){ int num=0,i; double m=-1,re; for(i=1;i<=n;i++){ re=dis(a,po[i]); if(re>m) m=re,num=i; } return num; } bool solve(double left,double right,double R){//三分列舉 double ans,mid; int num; while(left<=right){ mid=(left+right)/2; temp.x=cir.x+R*cos(mid); temp.y=cir.y+R*sin(mid); num=get_max_point(temp); ans=dis(temp,po[num]); if(ans<=r) return true; mid+=eps; temp.x=cir.x+R*cos(mid); temp.y=cir.y+R*sin(mid); mid-=eps; if(dis(temp,po[num]) > ans) right=mid-eps; else left=mid+eps; } return false; } int find_ans(){ double left=0,right=LL,ans=LL,mid,th,th1,a,b; int num,num1; point t,t1; while(left<=right){ mid=(left+right)/2; if(solve(0,PI,mid) || solve(PI,PI*2,mid)) ans=MIN(ans,mid),right=mid-eps; else left=mid+eps; } printf("%.2f\n",ans); return 0; } #define MAXN 2000 struct POINTSET{ double x, y; }; const double precison=1.0e-8; POINTSET maxcic, point[MAXN]; double radius; int curset[MAXN], posset[3]; int set_cnt, pos_cnt; inline double dis_2(POINTSET &from, POINTSET& to){ return ((from.x-to.x)*(from.x-to.x)+(from.y-to.y)*(from.y-to.y)); } int in_cic(int pt){ if(sqrt(dis_2(maxcic, point[pt]))<radius+precison) return 1; return 0; } int cal_mincic(){ if(pos_cnt==1 || pos_cnt==0) return 0; else if(pos_cnt==3){ double A1, B1, C1, A2, B2, C2; int t0=posset[0], t1=posset[1], t2=posset[2]; A1=2*(point[t1].x-point[t0].x); B1=2*(point[t1].y-point[t0].y); C1=point[t1].x*point[t1].x-point[t0].x*point[t0].x+ point[t1].y*point[t1].y-point[t0].y*point[t0].y; A2=2*(point[t2].x-point[t0].x); B2=2*(point[t2].y-point[t0].y); C2=point[t2].x*point[t2].x-point[t0].x*point[t0].x+ point[t2].y*point[t2].y-point[t0].y*point[t0].y; maxcic.y=(C1*A2-C2*A1)/(A2*B1-A1*B2); maxcic.x=(C1*B2-C2*B1)/(A1*B2-A2*B1); radius=sqrt(dis_2(maxcic, point[t0])); } else if(pos_cnt==2){ maxcic.x=(point[posset[0]].x+point[posset[1]].x)/2; maxcic.y=(point[posset[0]].y+point[posset[1]].y)/2; radius=sqrt(dis_2(point[posset[0]], point[posset[1]]))/2; } return 1; } int mindisk(){ if(set_cnt==0 || pos_cnt==3){ return cal_mincic(); } int tt=curset[--set_cnt]; int res=mindisk(); set_cnt++; if(!res || !in_cic(tt)){ set_cnt--; posset[pos_cnt++]=curset[set_cnt]; res=mindisk(); pos_cnt--; curset[set_cnt++]=curset[0]; curset[0]=tt; } return res; } int main1(int n){ int i; for(i=0; i<n; i++) point[i].x=po[i+1].x, point[i].y=po[i+1].y; if(n==1){ maxcic.x=point[0].x; maxcic.y=point[0].y; radius=0; } set_cnt=n; pos_cnt=0; for(i=0 ;i<n ;i++) curset[i]=i; mindisk(); return 0; } int main(){ int i,j,k; while(scanf("%lf%lf%lf",&cir.x,&cir.y,&r)!=EOF){ scanf("%d",&n); for(i=1;i<=n;i++) scanf("%lf%lf",&po[i].x,&po[i].y); if(n==1){ printf("%.2lf\n",dis(cir,po[1])-r); continue; } main1(n); if(radius>r){ printf("X\n");continue; } na.x=maxcic.x,na.y=maxcic.y; LL=dis(cir,na); find_ans(); } return 0; }