HDU 4667 Building Fence(求凸包的周長)
阿新 • • 發佈:2017-06-10
bsp algorithm building font stream def tmp using ios
題目鏈接:傳送門
題意:
給定n個圓。m個三角形求把這些圖形所有覆蓋的圖形的最小的周長。
分析:
剛開始看到就想到了求凸包,但是圓怎麽求了?就暴力把圓切割成1000個點然後求凸包就能夠了。
水過了。正解是找圓與圓的切點圓與三角形上的點與圓引得切線的切點。
代碼例如以下:
#include<stdio.h> #include<string.h> #include<algorithm> #include<cmath> #include<iostream> using namespace std; typedef long long LL; const double eps = 1e-10; const int maxn = 1e6+10; const double pi = acos(-1.0); const double inf = 1e17; int dcmp(double x){ if(fabs(x)<eps) return 0; if(x>0) return 1; return -1; } struct point{ double x,y; int id; point(){} point(double _x,double _y,int i):x(_x),y(_y),id(i){} bool operator <(const struct point &tmp)const{ if(dcmp(x-tmp.x)==0) return dcmp(y-tmp.y)<0; return dcmp(x-tmp.x)<0; } bool operator == (const struct point &tmp)const{ return dcmp(x-tmp.x)==0&&dcmp(y-tmp.y)==0; } }p[maxn],st[maxn]; double Cross(point a,point b,point c){ return (c.x-a.x)*(b.y-a.y) - (b.x-a.x)*(c.y-a.y); } int ConvexHull(point *p,int n,point *st){ sort(p,p+n); n=unique(p,p+n)-p; int m=0; for(int i=0; i<n; i++) { while(m>1&&Cross(st[m-2],p[i],st[m-1])<=0) m--; st[m++]=p[i]; } int k=m; for(int i=n-2; i>=0; i--){ while(m>k&&Cross(st[m-2],p[i],st[m-1])<=0)m--; st[m++]=p[i]; } if(n>1) m--; return m; } double dis(point a,point b){ return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); } int R[600]; int main() { int n,m; while(~scanf("%d%d",&n,&m)){ int cnt = 0; for(int i=0;i<n;i++){ double x,y,r; scanf("%lf%lf%lf",&x,&y,&r); R[i]=r; for(int j=0;j<1000;j++){ double tmp = 2*pi*j/1000; p[cnt++]=point(x+r*cos(tmp),y+r*sin(tmp),i); } } for(int i=0;i<m*3;i++){ double x,y; scanf("%lf%lf",&p[cnt].x,&p[cnt].y); p[cnt].id=cnt++; } int num = ConvexHull(p,cnt,st); double ans = 0; for(int i=0;i<num;i++){ int t1 = st[i].id; int t2 = st[(i+1)%num].id; if(t1==t2){ ans+=pi*2*R[t1]/1000; } else ans+=dis(st[i],st[(i+1)%num]); } printf("%.4lf\n",ans); } return 0; }
HDU 4667 Building Fence(求凸包的周長)