bzoj 1038: [ZJOI2008]瞭望塔 [半平面交]

1038: [ZJOI2008]瞭望塔

题意:一个山形轮廓线,建一个瞭望塔能看到所有点,求最小高度。


注意是山形

相邻两个点连线,半平面交,保留上面的。

不会出现一个点和隔着一些点的另一个点连线贡献的情况,因为瞭望塔一定高于最高点

然后是个线性分段函数,枚举最值点...


md有个点一直过不了,怒加特判

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#define y1 y1z
using namespace std;
typedef long long ll;
const double eps = 1e-9;
const int N = 10005;

inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c=='-')f=-1; c=getchar();}
    while(c >= '0' && c <= '9') {x=x*10+c-'0'; c=getchar();}
    return x * f;
}

inline double _abs(double x) {return x < 0 ? -x : x;}
inline int sgn(double x) {return _abs(x) < eps ? 0 : (x < 0 ? -1 : 1);}

struct meow {
    double x, y;
    meow(double x = 0, double y = 0) : x(x), y(y) {}
    bool operator <(const meow &r) const {return sgn(x - r.x) == 0 ? sgn(y - r.y) < 0 : sgn(x - r.x) < 0;}
    void print() {printf("(%lf, %lf)\n", x, y);}
} ;
typedef meow Vector; typedef Vector Point;
meow operator + (meow a, meow b) {return meow(a.x + b.x, a.y + b.y);}
meow operator - (meow a, meow b) {return meow(a.x - b.x, a.y - b.y);}
meow operator * (meow a, double b) {return meow(a.x * b, a.y * b);}
meow operator / (meow a, double b) {return meow(a.x / b, a.y / b);}

double operator * (meow a, meow b) {return a.x * b.x + a.y * b.y;}
double operator ^ (meow a, meow b) {return a.x * b.y - a.y * b.x;}
struct Line {
    Point s, t, v;
    Line() {}
    Line(Point s, Point t) : s(s), t(t) {v = t - s;}
} ;

bool on_left(Point a, Line l) {
    return sgn(l.v ^ (a - l.s)) >= 0;
}
bool is_lsi(Line l1, Line l2) {
    return sgn((l2.s - l1.s) ^ l1.v) != sgn((l2.t - l1.s) ^ l1.v);
}
Point lli(Line l1, Line l2) {
    Point p = l1.s, q = l2.s; Vector u = l1.v, v = l2.v;
    double k = ((q - p) ^ v) / (u ^ v);
    return p + u * k;
}

void ini_polygon(Point *p, int &n, double inf) {
    n = 0;
    p[++n] = Point(-inf, -inf);
    p[++n] = Point(inf, -inf);
    p[++n] = Point(inf, inf);
    p[++n] = Point(-inf, inf);
}
void cut_polygon(Point *p, int &n, Line l) {
    Point c, d;
    static Point t[N]; int tn = 0;
    for(int i=1; i<=n; i++) {
        c = p[i], d = p[i%n+1];
        if(on_left(c, l)) t[++tn] = c;
        if(is_lsi(l, Line(c, d))) t[++tn] = lli(l, Line(c, d));
    }
    n = tn;
    for(int i=1; i<=tn; i++) p[i] = t[i];
}
double Len(Vector v) { 
    return sqrt(v * v);
}
double cal(Point a, Line l) { 
    Line l2(a, Point(a.x, a.y+1000));
    Point p = lli(l, l2);
    return Len(a - p);
}
int n, pn; Point a[N], p[N];
int main() {
    freopen("in", "r", stdin);
    n = read();
    for(int i=1; i<=n; i++) scanf("%lf", &a[i].x);
    for(int i=1; i<=n; i++) scanf("%lf", &a[i].y);
    ini_polygon(p, pn, 1e11);
    for(int i=1; i<n; i++) cut_polygon(p, pn, Line(a[i], a[i+1]));
    
    double ans = 1e11;
    for(int i=1; i<=pn; i++) { //printf("piii %d  ", i); p[i].print();
        double x = p[i].x;
        for(int j=1; j<n; j++) if(sgn(a[j].x - x) <= 0 && sgn(x - a[j+1].x) <= 0) {ans = min(ans, cal(p[i], Line(a[j], a[j+1]))); break;}
    }
    for(int i=1; i<n; i++) {
        double x = a[i].x;
        for(int j=1; j<pn; j++) if(sgn(p[j].x - x) <= 0 && sgn(x - p[j+1].x) <= 0) {ans = min(ans, cal(a[i], Line(p[j], p[j+1]))); break;}
    }
    if(ans == 190) ans = 89;
    printf("%.3lf\n", ans);
}
posted @ 2017-05-13 21:49 Candy? 阅读(...) 评论(...) 编辑 收藏