HDU 4667 Building Fence

Question:

Give n circles and m triangles, and make sure they do not intersect each other. Use a fence to enclose them and find the shortest perimeter.

Practice: -- water over...

Cut a circle evenly into 500 points, and then calculate the convex hull.

Note: After the convex hull is obtained, remember to find the Special Section of the circle when calculating the perimeter.

Sin (x) x = PI * (I/j); // PI is used to represent the degree;

 

#include<stdio.h>#include<iostream>#include<math.h>#include<algorithm>#define PI  3.1415926using namespace std;struct list{    double x;    double y;    int r;    int pos;}point[500001],tu[500001],pp;int tops,num;int cmp1(struct list a,struct list b){    if(a.y!=b.y)return a.y<b.y;    else return a.x<b.x;}int cmp2(struct list a,struct list b){    if((a.x-point[0].x)*(b.y-point[0].y)==(b.x-point[0].x)*(a.y-point[0].y))        return ((a.x-point[0].x)*(a.x-point[0].x)+(a.y-point[0].y)*(a.y-point[0].y))<((point[0].x-b.x)*(point[0].x-b.x)+(point[0].y-b.y)*(point[0].y-b.y));    return (a.x-point[0].x)*(b.y-point[0].y)>(b.x-point[0].x)*(a.y-point[0].y);}int ts;void yuan(int x,int y,int r){    pp.r=r;    for(int i=0;i<500;i++)    {        double du;        du=(double)2.0*PI*i/500.0;        pp.x=1.0*x+1.0*r*cos(du);        pp.y=1.0*y+1.0*r*sin(du);        pp.pos=ts;        point[tops++]=pp;    }}int pan(int z,int b,int a){    double x1,y1,x2,y2;    if(a<0)return 1;    x1=tu[b].x-tu[a].x;    y1=tu[b].y-tu[a].y;    x2=point[z].x-tu[b].x;    y2=point[z].y-tu[b].y;    if(x1*y2>x2*y1)return 1;    return 0;}double len(struct list p1,struct list p2){    return (double)sqrt(1.0*((p1.x-p2.x)*(p1.x-p2.x)+1.0*(p1.y-p2.y)*(p1.y-p2.y)));}void init(int n,int m){    int i,j;    tops=0;    int a,b,c;    ts=0;    for(i=0;i<n;i++)    {        scanf("%d%d%d",&a,&b,&c);        ts++;        yuan(a,b,c);    }    for(i=0;i<m;i++)    for(j=0;j<3;j++)    {        scanf("%lf%lf",&point[tops].x,&point[tops].y);        point[tops].pos=0;        tops++;    }}void jiantu(){    int i;    sort(point,point+tops,cmp1);    sort(point+1,point+tops,cmp2);    tu[0]=point[0];    tu[1]=point[1];    tu[2]=point[2];    num=3;    for(i=3;i<tops;i++)    {        while(!pan(i,num-1,num-2))num--;        tu[num++]=point[i];    }}void print(){    double lens;    int i;    lens=0.0;    for(i=0;i<num;i++)    {        if(tu[i].pos>0&&(tu[i].pos==tu[(i+1)%num].pos))        {            lens+=1.0*tu[i].r*2*PI/500.0;        }        else        lens+=len(tu[i],tu[(i+1)%num]);    }    printf("%.5lf\n",lens);}int main(){    int n,m;   // freopen("1002.in","r",stdin);    while(~scanf("%d%d",&n,&m))    {        init(n,m);        jiantu();        print();    }    return 0;}