HDU 5033-B - Building-维护凸包-单调栈
2016-08-13 18:45
363 查看
http://acm.hdu.edu.cn/showproblem.php?pid=5033
题意:给一个数轴, 是地面。
地面上有n座楼,给出m个人,分别站在m个位置上,问每个位置能看见多大角度的天空(二维平面)
数据范围:
Input
The first line of the input contains an integer T, denoting the number of testcases. Then T test cases follow.
Each test case begins with a number N(1<=N<=10^5), the number of buildings.
In the following N lines, each line contains two numbers, x i(1<=x i<=10^7) and h i(1<=h i<=10^7).
After that, there's a number Q(1<=Q<=10^5) for the number of queries.
In the following Q lines, each line contains one number q i, which is the position Matt was at.
可以根据求凸包的过程求解,把人当成高度为0的楼插入数据中。
第一遍,按下标从左到右扫描,对于当前这个楼,如果与top楼形成的角度的绝对值,比楼top-1与楼top形成的角度绝对值还小(可用叉积判),就把top楼弹出,也就是,单调栈里维护的是角度绝对值增大的建筑群,这样能保证每个人看到的视野都是正确的。
同理,把下标反过来扫一遍,得到是右边的视野角度。
最后加起来就是答案
复杂度O(n)
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <iostream>
#include <queue>
#include <map>
#include <set>
#include <vector>
using namespace std;
const double PI=acos(-1.0);
struct node
{
double h;
double x;
int id;
node() {}
node (double a,double b)
{
x=a,h=b;
}
node operator - (const node &t)const
{
return node(x-t.x,h-t.h);
}
int operator * (const node &t)const /// 叉积 : 有向面积
{
return x*t.h-h*t.x;
}
};
node tm[100123*2];
int top;
double ans[100123*2];
node aa[100123*2];
bool cmp(node a,node b)
{
return a.x<b.x;
}
bool judge(const node& a,const node& b,const node &c)
{
return ((b-a)*(c-b))>=0;
}
double angle(const node &a, const node &b)
{
return atan((double)a.h / (double)(b.x - a.x));
}
int main()
{
int n,i;
int t;
cin>>t;
int cnt=1;
while(t--)
{
memset(ans,0,sizeof (ans));
cin>>n;
for (int i=1; i<=n; i++)
{
scanf("%lf%lf",&aa[i].x,&aa[i].h);
}
int cun=n;
int q;
cin>>q;
for (int i=1; i<= q; i++)
{
scanf("%lf",&aa[++cun].x);
aa[cun].h=0;
aa[cun].id=i;
}
sort(aa+1,aa+1+cun,cmp);
int top=0;
for (int i=1; i<=cun; i++)
{
while(top>=2 && judge (tm[top-1],tm[top],aa[i]))
top--;
if (aa[i].h>0)
tm[++top]=aa[i];
else
ans[aa[i].id]+=angle(tm[top],aa[i]);
}
reverse(aa+1, aa+1+cun);
top=0;
for (int i=1;i<=cun;i++)
aa[i].x=1e7-aa[i].x;
for (int i=1; i<=cun; i++)
{
while(top>=2 && judge (tm[top-1],tm[top],aa[i]))
top--;
if (aa[i].h>0)
tm[++top]=aa[i];
else
ans[aa[i].id]+=angle(tm[top],aa[i]);
}
printf("Case #%d:\n",cnt++);
for (int i=1;i<=q;i++)
printf("%.10lf\n", (PI - ans[i]) / PI * 180.0);//ans[i]存的是看不到的角度
}
return 0;
}
题意:给一个数轴, 是地面。
地面上有n座楼,给出m个人,分别站在m个位置上,问每个位置能看见多大角度的天空(二维平面)
数据范围:
Input
The first line of the input contains an integer T, denoting the number of testcases. Then T test cases follow.
Each test case begins with a number N(1<=N<=10^5), the number of buildings.
In the following N lines, each line contains two numbers, x i(1<=x i<=10^7) and h i(1<=h i<=10^7).
After that, there's a number Q(1<=Q<=10^5) for the number of queries.
In the following Q lines, each line contains one number q i, which is the position Matt was at.
可以根据求凸包的过程求解,把人当成高度为0的楼插入数据中。
第一遍,按下标从左到右扫描,对于当前这个楼,如果与top楼形成的角度的绝对值,比楼top-1与楼top形成的角度绝对值还小(可用叉积判),就把top楼弹出,也就是,单调栈里维护的是角度绝对值增大的建筑群,这样能保证每个人看到的视野都是正确的。
同理,把下标反过来扫一遍,得到是右边的视野角度。
最后加起来就是答案
复杂度O(n)
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <iostream>
#include <queue>
#include <map>
#include <set>
#include <vector>
using namespace std;
const double PI=acos(-1.0);
struct node
{
double h;
double x;
int id;
node() {}
node (double a,double b)
{
x=a,h=b;
}
node operator - (const node &t)const
{
return node(x-t.x,h-t.h);
}
int operator * (const node &t)const /// 叉积 : 有向面积
{
return x*t.h-h*t.x;
}
};
node tm[100123*2];
int top;
double ans[100123*2];
node aa[100123*2];
bool cmp(node a,node b)
{
return a.x<b.x;
}
bool judge(const node& a,const node& b,const node &c)
{
return ((b-a)*(c-b))>=0;
}
double angle(const node &a, const node &b)
{
return atan((double)a.h / (double)(b.x - a.x));
}
int main()
{
int n,i;
int t;
cin>>t;
int cnt=1;
while(t--)
{
memset(ans,0,sizeof (ans));
cin>>n;
for (int i=1; i<=n; i++)
{
scanf("%lf%lf",&aa[i].x,&aa[i].h);
}
int cun=n;
int q;
cin>>q;
for (int i=1; i<= q; i++)
{
scanf("%lf",&aa[++cun].x);
aa[cun].h=0;
aa[cun].id=i;
}
sort(aa+1,aa+1+cun,cmp);
int top=0;
for (int i=1; i<=cun; i++)
{
while(top>=2 && judge (tm[top-1],tm[top],aa[i]))
top--;
if (aa[i].h>0)
tm[++top]=aa[i];
else
ans[aa[i].id]+=angle(tm[top],aa[i]);
}
reverse(aa+1, aa+1+cun);
top=0;
for (int i=1;i<=cun;i++)
aa[i].x=1e7-aa[i].x;
for (int i=1; i<=cun; i++)
{
while(top>=2 && judge (tm[top-1],tm[top],aa[i]))
top--;
if (aa[i].h>0)
tm[++top]=aa[i];
else
ans[aa[i].id]+=angle(tm[top],aa[i]);
}
printf("Case #%d:\n",cnt++);
for (int i=1;i<=q;i++)
printf("%.10lf\n", (PI - ans[i]) / PI * 180.0);//ans[i]存的是看不到的角度
}
return 0;
}
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