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。

 

  1 #include<iostream>
  2 #include<algorithm>
  3 #include<cmath>
  4 #include<cstdio>
  5 #include<cstdlib>
  6 #include<cstring>
  7 using namespace std;
  8 
  9 #define eps (1e-6)
 10 #define oo ((double)(1ll<<50))
 11 #define maxn 310
 12 int n,m,tot,cnt;
 13 double ans = oo;
 14 struct NODE
 15 {
 16     double x,y;
 17     friend inline NODE operator + (const NODE &p,const NODE &q) { return (NODE) {p.x+q.x,p.y+q.y}; }
 18     friend inline NODE operator - (const NODE &p,const NODE &q) { return (NODE) {p.x-q.x,p.y-q.y}; }
 19     friend inline NODE operator * (const NODE &p,const double &q) { return (NODE) {p.x*q,p.y*q}; }
 20     friend inline double operator /(const NODE &p,const NODE &q) { return p.x*q.y-p.y*q.x; }
 21     inline double alpha() { return atan2(y,x); }
 22 }mou[maxn],pol[maxn],pp[maxn];
 23 struct LINE
 24 {
 25     NODE p,v; double slop;
 26     inline void maintain() { slop = v.alpha(); }
 27     friend inline bool operator <(const LINE &l1,const LINE &l2) { return l1.slop < l2.slop; }
 28 }lines[maxn],qq[maxn];
 29 struct SCAN
 30 {
 31     double x,y; int id; bool sign;
 32     friend inline bool operator <(const SCAN &a,const SCAN &b)
 33     {
 34         if (a.x != b.x) return a.x < b.x;
 35         else return a.sign < b.sign;
 36     }
 37 }bac[maxn];
 38 
 39 inline bool ol(const LINE &l,const NODE &p) { return l.v/(p-l.p) > 0; }
 40 
 41 inline NODE cp(const LINE &a,const LINE &b)
 42 {
 43     NODE u = a.p - b.p;
 44     double t = (b.v/u)/(a.v/b.v);
 45     return a.p+a.v*t;
 46 }
 47 
 48 inline bool para(const LINE &a,const LINE &b)
 49 {
 50     return fabs(a.v/b.v) < eps;
 51 }
 52 
 53 inline void ready()
 54 {
 55     for (int i = 1;i < n;++i)
 56     {
 57         lines[++tot] = (LINE) {mou[i],(mou[i+1]-mou[i])*1e-3};
 58         lines[tot].maintain();
 59     }
 60     lines[++tot] = (LINE) {(NODE) {-oo,0},(NODE){0,-0.001}};
 61     lines[tot].maintain();
 62     
 63     lines[++tot] = (LINE) {(NODE) {0,oo},(NODE){-0.001,0}};
 64     lines[tot].maintain();
 65     
 66     lines[++tot] = (LINE) {(NODE) {oo,0},(NODE){0,0.001}};
 67     lines[tot].maintain();
 68     
 69     lines[++tot] = (LINE) {(NODE) {0,-oo},(NODE){0.001,0}};
 70     lines[tot].maintain();
 71 }
 72 
 73 inline int half_plane_intersection()
 74 {
 75     sort(lines+1,lines+tot+1);
 76     int head,tail;
 77     qq[head = tail = 1] = lines[1];
 78     for (int i = 2;i <= tot;++i)
 79     {
 80         while (head < tail&&!ol(lines[i],pp[tail-1])) --tail;
 81         while (head < tail&&!ol(lines[i],pp[head])) ++head;
 82         qq[++tail] = lines[i];
 83         if (para(qq[tail],qq[tail-1]))
 84         {
 85             tail--;
 86             if (ol(qq[tail],lines[i].p)) qq[tail] = lines[i];
 87         }
 88         if (head < tail) pp[tail-1] = cp(qq[tail],qq[tail-1]);
 89     }
 90     while (head < tail && !ol(qq[head],pp[tail-1])) --tail;
 91     if (tail-head <= 0) return 0;
 92     pp[tail] = cp(qq[tail],qq[head]);
 93     for (int i = head;i <= tail;++i) pol[++m] = pp[i];
 94     pol[0] = pol[m];
 95     return m;
 96 }
 97 
 98 inline void work()
 99 {
100     int all = 0;
101     for (int i = 1;i <= n;++i)
102         bac[++all] = (SCAN) { mou[i].x,mou[i].y,i,false };
103     for (int i = 1;i <= m;++i)
104         if (pol[i].x >= mou[1].x&&pol[i].x <= mou[n].x)
105             bac[++all] = (SCAN) { pol[i].x,pol[i].y,i,true };
106     sort(bac+1,bac+all+1);
107     int s1,s2;
108     for (int i = 1;i <= all;++i) if (bac[i].sign) { s1 = bac[i].id-1; break; }
109     for (int i = 1;i <= all;++i)
110     {
111         LINE l = (LINE) {(NODE) {bac[i].x,0},(NODE) {0,1}},l1; NODE p;
112         if (!bac[i].sign)
113         {
114             l1= (LINE) {pol[s1],pol[s1+1]-pol[s1]};
115             s2 = bac[i].id;
116         }
117         else
118         {
119             l1= (LINE) {mou[s2],mou[s2+1]-mou[s2]};
120             s1 = bac[i].id;
121         }
122         p = cp(l,l1);
123         ans = min(ans,fabs(p.y-bac[i].y));    
124     }
125  }
126 
127 int main()
128 {
129     freopen("1038.in","r",stdin);
130     freopen("1038.out","w",stdout);
131     scanf("%d ",&n);
132     for (int i = 1;i <= n;++i) scanf("%lf",&mou[i].x);
133     for (int i = 1;i <= n;++i) scanf("%lf",&mou[i].y);
134     ready();
135     half_plane_intersection();
136     work();
137     printf("%.3lf",ans);
138     fclose(stdin); fclose(stdout);
139     return 0;
140 }
View Code

 

高考结束,重新回归。
posted @ 2015-02-05 22:29  lmxyy  阅读(160)  评论(0编辑  收藏  举报