wf 2017A

给一个多边形，问能放进去的最长的线段的长度。

我调了两个小时结果加了inline就过？？？什么东西啊。2000+MS->890MS

真实自闭啊。

dls寒假已经讲的很清楚了(别问我为什么现在才做)

就是枚举所有点对然后  求出来这条直线 与多边形所有点的 交点，然后遍历这些交点，相当于进行一个最大字段和的操作。

如何check线段是不是在内部的话取线段中点就可以

调了一晚上简直自闭了，我寻思我剪枝还比别人多，做法一样的怎么我就过不去呢？？

感觉把campdiv2消化干净的话可以get到超级多的新姿势呢

 #include <cstdio>
 #include <algorithm>
 #include <cmath>
 #include <vector>
 using namespace std;
 typedef double db;
 const db eps = 1e-;
 const db pi = acos(-);
 inline int sign(db k){
     if (k>eps) return ; else if (k<-eps) return -; return ;
 }
 inline int cmp(db k1,db k2){return sign(k1-k2);}
 inline int inmid(db k1,db k2,db k3){return sign(k1-k3)*sign(k2-k3)<=;}// k3 在 [k1,k2] 内
 struct point{
     db x,y;
     point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};}
     point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};}
     point operator * (db k1) const{return (point){x*k1,y*k1};}
     point operator / (db k1) const{return (point){x/k1,y/k1};}
     int operator == (const point &k1) const{return cmp(x,k1.x)==&&cmp(y,k1.y)==;}
     bool operator<(const point &k1)const {
         int c = cmp(x,k1.x);
         if(c)return c==-;
         return cmp(y,k1.y)==-;
     }
     inline db abs(){ return sqrt(x*x+y*y);}
     inline db abs2(){return x*x+y*y;}
     inline db dis(point k1){return (*this-k1).abs();}
     int getP() const{return sign(y)==||(sign(y)==&&sign(x)==-);}
 };
 inline db cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;}
 inline db dot(point k1,point k2){return k1.x*k2.x+k1.y*k2.y;}
 inline int inmid(point k1,point k2,point k3){return inmid(k1.x,k2.x,k3.x)&&inmid(k1.y,k2.y,k3.y);}
 inline int onS(point k1,point k2,point q){return inmid(k1,k2,q)&&sign(cross(k1-q,k2-k1))==;}
 point proj(point k1,point k2,point q){
     point k=k2-k1;return k1+k*(dot(q-k1,k)/k.abs2());
 }
 inline int checkLL(point k1,point k2,point k3,point k4){//判重合或者平行
     return cmp(cross(k3-k1,k4-k1),cross(k3-k2,k4-k2))!=;
 }
 point getLL(point k1,point k2,point k3,point k4){
     db w1 = cross(k1-k3,k4-k3),w2=cross(k4-k3,k2-k3);
     return (k1*w2+k2*w1)/(w1+w2);
 }
 int n;point p[];
 inline int contain(point q){ // 2 内部 1 边界 0 外部
     int pd=;
     for (int i=;i<=n;i++){
         point u=p[i-],v=p[i];
         if (onS(u,v,q)) return ; if (cmp(u.y,v.y)>) swap(u,v);
         if (cmp(u.y,q.y)>=||cmp(v.y,q.y)<) continue;
         if (sign(cross(u-v,q-v))<) pd^=;
     }
     return pd<<;
 }
 db ans=;
 inline void slove(point x,point y){
     vector<point> v;
     for(int i=;i<n;i++){
         if(sign(cross(y-x,p[i]-x)*cross(y-x,p[i+]-x))<= ){
             point v1 = y-x,v2=p[i+]-p[i];
             if(sign(cross(v1,v2)) == ){
                 v.push_back(p[i]);
                 v.push_back(p[i+]);
             }
             else v.push_back(getLL(x,y,p[i],p[i+]));
         }
     }
     sort(v.begin(),v.end());
     v.resize(unique(v.begin(),v.end())-v.begin());
     db tmp=;int m = v.size()-;
     for(int i=;i<m;i++){
         point mid = (v[i]+v[i+])/;
         if(contain(mid)){
             tmp+=v[i+].dis(v[i]);
         }else{
             ans = max(ans,tmp);
             tmp = ;
             if(v[i+].dis(v[m])<=ans)
                 return;
         }
     }
     ans = max(ans,tmp);
 }
 int main(){
     //freopen("secret-046.in","r",stdin);
     scanf("%d",&n);
     for(int i=;i<n;i++){
         scanf("%lf%lf",&p[i].x,&p[i].y);
     }
     p[n]=p[];
     for(int i=;i<n-;i++){
         for(int j=i+;j<n;j++){
             slove(p[i],p[j]);
         }
     }
     printf("%.9f\n",ans);
 }
 /**
 5
 2 0
 2 3
 1 1
 0 2
 0 0
  */