就是……判断p与圆心的连线与圆的交点在不在圆弧上,在就是它到p的距离,不在就是p跟端点的最短距离
#include<iostream>
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
const double pi=acos(-1.0);
struct dot
{
double x,y;
dot(){}
dot(double a,double b){x=a;y=b;}
dot operator +(dot a){return dot(x+a.x,y+a.y);}
......
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