这一题WA了好久，对拍才查出来错，因为放入单调栈的时候少考虑了前两种元素(1,10),(2,13)这种case，忘了把第二个放进去。
维护一个单调栈(凸包)，两个方向各遍历一次。每次只遍历单调栈，可构成凸包截止。
#include<iostream>
#include<stdio.h>
#include<cstdio>
#include<stdlib.h>
#include<vector>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<stack>
#include<queue>
#include <ctype.h>
using namespace std;
int T;
int N;
int Q;
const int maxn=200100;
pair<int,int>a[maxn];//first is x,second is h
pair<int,int> s[maxn];
int top;//top of stack s,指向下一个代存位置
pair<int,int> l[maxn];
pair<int,int> r[maxn];
int q[maxn];//record the location of query
const double pi=acos(-1.0);
bool cmp(pair<int,int>a,pair<int,int>b)
{
return a.first<b.first;
}
void input()
{
memset(a,0,sizeof(a));
memset(l,0,sizeof(l));
memset(r,0,sizeof(r));
memset(q,0,sizeof(q));
scanf("%d",&N);
for(int i=0;i<N;i++)
{
scanf("%d %d",&a[i].first,&a[i].second);
}
scanf("%d",&Q);
for(int i=N;i<N+Q;i++)
{
scanf("%d",&a[i].first);
q[i-N]=a[i].first;
a[i].second=-(i-N+1);// if<0, then it represent the order of query
}
sort(a,a+N+Q,cmp);
// for(int i=0;i<N+Q;i++)
// {
// cout<<a[i].first<<" "<<a[i].second<<endl;
// }
}
void convex(int x)
{
pair<int,int>tmp=s[top];
// cout<<top<<endl;
while(top>0)
{
tmp=s[top-1];//the first element
if(tmp.second<a[x].second)//need =?
{
top--;//to keep descending
}
else
{
break;
}
}
// s[top++]=a[x];
while(top>=2)//at least two elements
{
tmp=s[top-1];
double slope1=1.0*(s[top-2].second-tmp.second)/abs(tmp.first-s[top-2].first);
double slope2=1.0*(tmp.second-a[x].second)/abs(a[x].first-tmp.first);
if(slope1>=slope2)
{
top--;
}
else
{
s[top++]=a[x];
// cout<<x<<" xx "<<tmp.second<<" "<<a[x].second<<endl;
// cout<<x<<" "<<s[top-1].first<<endl;
return;
}
}
if(top==1||top==0)
{
s[top++]=a[x];
// cout<<x<<"xx "<<s[top-1].second<<endl;
// cout<<x<<" "<<s[top-1].first<<endl;
return;
}
}
void run()
{
memset(s,0,sizeof(s));
// memset(l,0,sizeof(l));
// memset(r,0,sizeof(r));
top=0;
s[top++]=a[0];
for(int i=1;i<N+Q;i++)
{
//cout<<s[top-1].first<<endl;
if(a[i].second>0)
{
convex(i);
}
else
{
while(top>=2)
{
pair<int,int>tmp=s[top-1];
double slope1=1.0*(s[top-2].second-tmp.second)/abs(tmp.first-s[top-2].first);
double slope2=1.0*tmp.second/abs(a[i].first-tmp.first);
// cout<<slope1<<" "<<slope2<<endl;
if(slope1>=slope2)
{
top--;
}
else
{
break;
}
}
int qry=-a[i].second-1;
l[qry]=s[top-1];
// cout<<l[qry].first<<endl;;
}
}
memset(s,0,sizeof(s));//from right end
// cout<<"right"<<endl;
top=0;
s[top++]=a[N+Q-1];
for(int i=N+Q-2;i>=0;i--)
{
if(a[i].second>0)
{
convex(i);
//cout<<s[top-1].first<<endl;
}
else
{
while(top>=2)
{
pair<int,int>tmp=s[top-1];
double slope1=1.0*(s[top-2].second-tmp.second)/abs(tmp.first-s[top-2].first);
double slope2=1.0*tmp.second/abs(a[i].first-tmp.first);
// cout<<tmp.first<<" tmp "<<tmp.second<<endl;
if(slope1>=slope2)
{
top--;
}
else
{
break;
}
}
int qry=-a[i].second-1;
r[qry]=s[top-1];
// cout<<top-1<<" "<<qry<<" "<<r[qry].first<<endl;
}
}
}
double cal(int x)
{
// cout<<x<<" "<<l[x].second<<" "<<q[x]<<" "<<l[x].first<<endl;
// cout<<x<<" "<<r[x].second<<" "<<q[x]<<" "<<r[x].first<<endl;
double theta1=atan2(l[x].second,(q[x]-l[x].first));
double theta2=atan2(r[x].second,(r[x].first-q[x]));
return (pi-theta1-theta2)/pi*180.0;
}
int main()
{
// freopen("input.txt","r",stdin);
freopen("data.txt","r",stdin);
// freopen("out1.txt","w",stdout);
scanf("%d",&T);
for(int ca=1;ca<=T;ca++)
{
input();
run();
printf("Case #%d:\n",ca);
for(int i=0;i<Q;i++)
{
printf("%.10lf\n",cal(i));
}
}
return 0;
}