BZOJ 1038 瞭望塔

Description

致力于建设全国示范和谐小村庄的H村村长dadzhi，决定在村中建立一个瞭望塔，以此加强村中的治安。我们将H村抽象为一维的轮廓。如下图所示 我们可以用一条山的上方轮廓折线(x1, y1), (x2, y2), …. (xn, yn)来描述H村的形状，这里x1 < x2 < …< xn。瞭望塔可以建造在[x1, xn]间的任意位置, 但必须满足从瞭望塔的顶端可以看到H村的任意位置。可见在不同的位置建造瞭望塔，所需要建造的高度是不同的。为了节省开支，dadzhi村长希望建造的塔高度尽可能小。请你写一个程序，帮助dadzhi村长计算塔的最小高度。

Input

第一行包含一个整数n，表示轮廓折线的节点数目。接下来第一行n个整数, 为x1 ~ xn. 第三行n个整数，为y1 ~ yn。

Output

仅包含一个实数，为塔的最小高度，精确到小数点后三位。

Sample Input

【输入样例一】
6
1 2 4 5 6 7
1 2 2 4 2 1
【输入样例二】
4
10 20 49 59
0 10 10 0

Sample Output

【输出样例一】
1.000
【输出样例二】
14.500

HINT

对于100%的数据， N ≤ 300，输入坐标绝对值不超过106，注意考虑实数误差带来的问题。

Source

半平面交。对于每条线段，所能看到其整条线段的点一定的在其所延长直线的上方，因此我们可以对所以直线求一次半平面交。

然后，最优解一定在线段端点处或半平面交所得多边形的顶点处。

 #include<iostream>
 #include<algorithm>
 #include<cmath>
 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 using namespace std;

 #define eps (1e-6)
 #define oo ((double)(1ll<<50))
 #define maxn 310
 int n,m,tot,cnt;
 double ans = oo;
 struct NODE
 {
     double x,y;
     friend inline NODE operator + (const NODE &p,const NODE &q) { return (NODE) {p.x+q.x,p.y+q.y}; }
     friend inline NODE operator - (const NODE &p,const NODE &q) { return (NODE) {p.x-q.x,p.y-q.y}; }
     friend inline NODE operator * (const NODE &p,const double &q) { return (NODE) {p.x*q,p.y*q}; }
     friend inline double operator /(const NODE &p,const NODE &q) { return p.x*q.y-p.y*q.x; }
     inline double alpha() { return atan2(y,x); }
 }mou[maxn],pol[maxn],pp[maxn];
 struct LINE
 {
     NODE p,v; double slop;
     inline void maintain() { slop = v.alpha(); }
     friend inline bool operator <(const LINE &l1,const LINE &l2) { return l1.slop < l2.slop; }
 }lines[maxn],qq[maxn];
 struct SCAN
 {
     double x,y; int id; bool sign;
     friend inline bool operator <(const SCAN &a,const SCAN &b)
     {
         if (a.x != b.x) return a.x < b.x;
         else return a.sign < b.sign;
     }
 }bac[maxn];

 inline bool ol(const LINE &l,const NODE &p) { return l.v/(p-l.p) > ; }

 inline NODE cp(const LINE &a,const LINE &b)
 {
     NODE u = a.p - b.p;
     double t = (b.v/u)/(a.v/b.v);
     return a.p+a.v*t;
 }

 inline bool para(const LINE &a,const LINE &b)
 {
     return fabs(a.v/b.v) < eps;
 }

 inline void ready()
 {
     for (int i = ;i < n;++i)
     {
         lines[++tot] = (LINE) {mou[i],(mou[i+]-mou[i])*1e-};
         lines[tot].maintain();
     }
     lines[++tot] = (LINE) {(NODE) {-oo,},(NODE){,-0.001}};
     lines[tot].maintain();

     lines[++tot] = (LINE) {(NODE) {,oo},(NODE){-0.001,}};
     lines[tot].maintain();

     lines[++tot] = (LINE) {(NODE) {oo,},(NODE){,0.001}};
     lines[tot].maintain();

     lines[++tot] = (LINE) {(NODE) {,-oo},(NODE){0.001,}};
     lines[tot].maintain();
 }

 inline int half_plane_intersection()
 {
     sort(lines+,lines+tot+);
     int head,tail;
     qq[head = tail = ] = lines[];
     for (int i = ;i <= tot;++i)
     {
         while (head < tail&&!ol(lines[i],pp[tail-])) --tail;
         while (head < tail&&!ol(lines[i],pp[head])) ++head;
         qq[++tail] = lines[i];
         if (para(qq[tail],qq[tail-]))
         {
             tail--;
             if (ol(qq[tail],lines[i].p)) qq[tail] = lines[i];
         }
         if (head < tail) pp[tail-] = cp(qq[tail],qq[tail-]);
     }
     while (head < tail && !ol(qq[head],pp[tail-])) --tail;
     if (tail-head <= ) return ;
     pp[tail] = cp(qq[tail],qq[head]);
     for (int i = head;i <= tail;++i) pol[++m] = pp[i];
     pol[] = pol[m];
     return m;
 }

 inline void work()
 {
     int all = ;
     for (int i = ;i <= n;++i)
         bac[++all] = (SCAN) { mou[i].x,mou[i].y,i,false };
     for (int i = ;i <= m;++i)
         if (pol[i].x >= mou[].x&&pol[i].x <= mou[n].x)
             bac[++all] = (SCAN) { pol[i].x,pol[i].y,i,true };
     sort(bac+,bac+all+);
     int s1,s2;
     for (int i = ;i <= all;++i) if (bac[i].sign) { s1 = bac[i].id-; break; }
     for (int i = ;i <= all;++i)
     {
         LINE l = (LINE) {(NODE) {bac[i].x,},(NODE) {,}},l1; NODE p;
         if (!bac[i].sign)
         {
             l1= (LINE) {pol[s1],pol[s1+]-pol[s1]};
             s2 = bac[i].id;
         }
         else
         {
             l1= (LINE) {mou[s2],mou[s2+]-mou[s2]};
             s1 = bac[i].id;
         }
         p = cp(l,l1);
         ans = min(ans,fabs(p.y-bac[i].y));
     }
  }

 int main()
 {
     freopen("1038.in","r",stdin);
     freopen("1038.out","w",stdout);
     scanf("%d ",&n);
     for (int i = ;i <= n;++i) scanf("%lf",&mou[i].x);
     for (int i = ;i <= n;++i) scanf("%lf",&mou[i].y);
     ready();
     half_plane_intersection();
     work();
     printf("%.3lf",ans);
     fclose(stdin); fclose(stdout);
     return ;
 }