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  • Most Distant Point from the Sea UVALive

    Most Distant Point from the Sea

     UVALive - 3890 

    题意:给n个点(凸包),让求距离边界最远的点到边界的距离.

    二分答案, 若所有边的半平面交不为空则有解.

      1 /*************************************************************************
      2     > File Name: board.cpp
      3     > Author: yijiull
      4     > Mail: 1147161372@qq.com 
      5     > Created Time: 2017年09月22日 星期五 21时19分51秒
      6  ************************************************************************/
      7 #include <iostream>
      8 #include <cstring>
      9 #include <cstdio>
     10 #include <bits/stdc++.h>
     11 using namespace std;
     12 #define FP freopen("in.txt", "r", stdin)
     13 const double eps = 1e-12;
     14 const int inf = 0x3f3f3f3f;
     15 struct Point {
     16     double x,y;
     17     Point (double x = 0, double y = 0) : x(x), y(y) {}
     18 };
     19 typedef Point Vector;
     20 Vector operator + (Vector a, Vector b) {
     21     return Vector (a.x + b.x, a.y + b.y);
     22 }
     23 Vector operator * (Vector a, double s) {
     24     return Vector (a.x * s, a.y * s);
     25 }
     26 Vector operator / (Vector a, double p) {
     27     return Vector (a.x / p, a.y / p);
     28 }
     29 Vector operator - (Point a, Point b) {
     30     return Vector (a.x - b.x, a.y - b.y);
     31 }
     32 bool operator < (Point a, Point b) {
     33     return a.x < b.x || (a.x == b.x && a.y < b.y);
     34 }
     35 int dcmp (double x) {
     36     if(fabs(x) < eps) return 0;
     37     return x < 0 ? -1 : 1;
     38 }
     39 bool operator == (const Point &a, const Point &b) {
     40     return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
     41 }
     42 double Dot(Vector a, Vector b) {
     43     return a.x * b.x + a.y * b.y;
     44 }
     45 double Length (Vector a) {
     46     return sqrt(Dot(a, a));
     47 }
     48 double Cross (Vector a, Vector b) {
     49     return a.x * b.y - a.y * b.x;
     50 }
     51 Vector Normal (Vector a) {
     52     double L = Length(a);
     53     return Vector(-a.y / L, a.x / L);
     54 }
     55 //半平面交
     56 struct Line{
     57     Point p;
     58     Vector v;
     59     double rad;
     60     Line () {}
     61     Line (Point p, Vector v) : p(p), v(v) {
     62         rad = atan2(v.y,v.x);
     63     }
     64     bool operator < (const Line &L) const {
     65         return rad < L.rad;
     66     }
     67 };
     68 bool OnLeft(Line L, Point p) {
     69     return Cross(L.v, p - L.p) > 0;
     70 }
     71 Point GetLineIntersection (Line a, Line b) {
     72     Vector u = a.p - b.p;
     73     double t = Cross(b.v, u) / Cross(a.v, b.v);
     74     return a.p + a.v*t;
     75 }
     76 int HalfplaneIntersection(Line *L, int n, Point *poly) {
     77     sort(L, L+n);
     78     int first,last;
     79     Point *p = new Point[n];
     80     Line *q = new Line[n];  //双端队列
     81     q[first = last = 0] = L[0];
     82     for(int i = 1; i < n; i++) {
     83         while(first < last && !OnLeft(L[i], p[last-1])) last--;   //去尾
     84         while(first < last && !OnLeft(L[i], p[first])) first++; 
     85         q[++last] = L[i];
     86         if(dcmp(Cross(q[last].v, q[last-1].v)) == 0) {
     87             last--;
     88             if(OnLeft(q[last], L[i].p)) q[last] = L[i];
     89         }
     90         if(first < last) p[last-1] = GetLineIntersection (q[last-1],q[last]);
     91     }
     92     while(first < last && !OnLeft(q[first], p[last-1])) last--;  //删除无用平面
     93     if(last - first <= 1) return 0;  //空集
     94     p[last] = GetLineIntersection (q[last], q[first]);
     95     int m = 0;
     96     for(int i = first; i <= last; i++) poly[m++] = p[i];
     97     return m;
     98 }
     99 Point p[200], poly[200];
    100 Line L[200];
    101 Vector v[200], v2[200];
    102 int main(){
    103     int n;
    104     //FP;
    105     while(scanf("%d", &n) && n) {
    106         int m, x, y;
    107         for(int i = 0; i < n; i++) {
    108             scanf("%d%d", &x, &y);
    109             p[i] = Point(x,y);
    110         }
    111         for(int i = 0; i < n; i++) {
    112             v[i] = p[(i+1)%n] - p[i];
    113             v2[i] = Normal(v[i]);
    114         }
    115         double Ls = 0, Rs = 20000;
    116         while(Rs - Ls > 1e-6) {
    117             double mid = Ls + (Rs - Ls)/2;
    118             for(int i = 0; i < n; i++) L[i] = Line(p[i] + v2[i]*mid, v[i]);
    119             m = HalfplaneIntersection (L, n, poly);
    120             if(!m) Rs = mid;
    121             else Ls = mid;
    122         }
    123         printf("%.6lf
    ", Ls);
    124     }
    125     return 0;
    126 }
    View Code
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  • 原文地址:https://www.cnblogs.com/yijiull/p/7577589.html
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