hdu 4667（凸包）

<span style="font-family:FangSong_GB2312;font-size:18px;">#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <string>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <functional>

using namespace std;
const int MAXN=100005;
const double eps=1e-10;
const double pi=acos(-1.0);
int dcmp(double x){
if(fabs(x)<eps)return 0;
if(x>0)return 1;
return -1;
}                                                   //因为精度的问题，在这里求减法的精度
struct Point {
double x,y;
int id;                                        //所在的图形
}p[MAXN];                                          //几何图形中的点集
double dot(Point a,Point b,Point c){
double s1=b.x-a.x;
double t1=b.y-a.y;
double s2=c.x-a.x;
double t2=c.y-a.y;
return s1*s2+t1*t2;
}                                                 //求解点积
int n,res[MAXN],top;
bool cmp(Point a,Point b){
if(a.y==b.y)return a.x<b.x;
return a.y<b.y;
}                                                  //凸包的排序
bool mult(Point sp,Point ep,Point op){
return (sp.x-op.x)*(ep.y-op.y)>=(ep.x-op.x)*(sp.y-op.y);
}                                                  //凸包中的判断
void Graham(){
int len;
top=1;
sort(p,p+n,cmp);
if(n==0)return;res[0]=0;
if(n==1)return;res[1]=1;
if(n==2)return;res[2]=2;
for(int i=2;i<n;i++){
while(top&&mult(p[i],p[res[top]],p[res[top-1]]))top--;
res[++top]=i;
}
len=top;
res[++top]=n-2;
for(int i=n-3;i>=0;i--){
while(top!=len&&mult(p[i],p[res[top]],p[res[top-1]]))top--;
res[++top]=i;
}
}                                                    //求解凸包
double dis(Point a,Point b){
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}                                                    //两点间距离
double R[1000];
int main()
{
int x,m;
while(scanf("%d%d",&x,&m)!=EOF){
n=0;
double x1,y1,r;
for(int i=0;i<x;i++){
scanf("%lf%lf%lf",&x1,&y1,&r);
R[i]=r;
for(int j=0;j<1000;j++){
double tmp=2*pi*j/1000;            //分解圆
p
.id=i;
p
.x=x1+r*cos(tmp);
p[n++].y=y1+r*sin(tmp);
}
}
while(m--){
for(int i=1;i<=6;i++){
scanf("%lf",&r);
if(i&1)p
.x=r;
else{
p
.id=n;
p[n++].y=r;
}
}
}
Graham();
double ans=0;
for(int i=0;i<top;i++){
int t1=p[res[i]].id;
int t2=p[res[(i+1)%top]].id;
if(t1==t2){
ans+=pi*2*R[t1]/1000;
}
else ans+=dis(p[res[i]],p[res[(i+1)%top]]);        //求和
}
printf("%.4lf\n",ans);
}
return 0;
}</span>