hdu5033 Building 北京网络赛B题 单调栈

Problem Description

Once upon a time Matt went to a small town. The town was so small and narrow that he can regard the town as a pivot. There were some skyscrapers in the town, each located at position x _i with its height h _i. All skyscrapers located in different place. The skyscrapers had no width, to make it simple. As the skyscrapers were so high, Matt could hardly see the sky.Given the position Matt was at, he wanted to know how large the angle range was where he could see the sky. Assume that Matt's height is 0. It's guaranteed that for each query, there is at least one building on both Matt's left and right, and no building locate at his position.

 

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.

 

Output

For each test case, first output one line "Case #x:", where x is the case number (starting from 1).

Then for each query, you should output the angle range Matt could see the sky in degrees. The relative error of the answer should be no more than 10^(-4).

 

Sample Input

   
   
   
   

    
    
    
    3
3
1 2
2 1
5 1
1
4
3
1 3
2 2
5 1
1
4
3
1 4
2 3
5 1
1
4
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case #1:
101.3099324740
Case #2:
90.0000000000
Case #3:
78.6900675260
   
   
   
   

  给出一些房子的坐标和高度，询问一些坐标，输出这个坐标能看见天空的角度。

  把询问点当成坐标为x高度为0的房子，和别的房子一起按x坐标排序。先从左到右扫一遍，栈中存斜率不增的点，也就是发现K(stk[rear-1],stk[rear])<K(stk[rear],i)就退栈，再把i这个点加入栈中，并记录栈顶元素在左边挡住了i。因为被退掉的点是肯定不会挡住视线的。同理从右到左扫一遍，栈中存斜率不减的点，同样的方法记住右边挡住i的是第几个房子。最后对于每个询问，根据左边挡住它的房子和右边挡住它的房子计算角度。

#include<iostream>
#include<algorithm>
#include<queue>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<set>
#include<map>
#define INF 0x3f3f3f3f
#define MAXN 200010
#define MAXM 110
#define MOD 1000000007
#define MAXNODE 4*MAXN
#define eps 1e-9
#define pi (4*atan(1.0))
using namespace std;
typedef long long LL;
int T,N,Q,stk[MAXN];
struct Point{
    double x,h,ans;
    int l,r,id;
    bool operator < (const Point& rhs) const{
        return id<rhs.id;
    }
}p[MAXN];
bool cmp(Point a,Point b){
    return a.x<b.x;
}
double K(int a,int b){
    return (p[a].h-p[b].h)/(p[a].x-p[b].x);
}
void solve1(){
    int rear=0;
    stk[0]=0;
    for(int i=1;i<N;i++){
        while(rear&&K(stk[rear-1],stk[rear])<K(stk[rear],i)) rear--;
        p[i].l=p[stk[rear]].id;
        stk[++rear]=i;
    }
}
void solve2(){
    int rear=0;
    stk[0]=N-1;
    for(int i=N-2;i>=0;i--){
        while(rear&&K(stk[rear-1],stk[rear])>K(stk[rear],i)) rear--;
        p[i].r=p[stk[rear]].id;
        stk[++rear]=i;
    }
}
double cal(double x1,double y1,double x2,double y2){
    return 180*(atan2(y1,x1)-atan2(y2,x2))/pi;
}
int main(){
    freopen("in.txt","r",stdin);
    scanf("%d",&T);
    int cas=0;
    while(T--){
        p[0].h=p[0].id=0;
        p[0].x=-1;
        scanf("%d",&N);
        int n=N;
        for(int i=1;i<=N;i++){
            scanf("%lf%lf",&p[i].x,&p[i].h);
            p[i].id=i;
        }
        scanf("%d",&Q);
        for(int i=1;i<=Q;i++){
            scanf("%lf",&p[N+i].x);
            p[N+i].h=0;
            p[N+i].id=N+i;
        }
        p[N+Q+1].x=10000000000;
        p[N+Q+1].h=0;
        p[N+Q+1].id=N+Q+1;
        N=N+Q+2;
        sort(p,p+N,cmp);
        solve1();
        solve2();
        sort(p,p+N);
        printf("Case #%d:\n",++cas);
        for(int i=n+1;i<=n+Q;i++) printf("%.10lf\n",cal(p[p[i].l].x-p[i].x,p[p[i].l].h,p[p[i].r].x-p[i].x,p[p[i].r].h));
    }
    return 0;
}