Description
The main land of Japan called Honshu is an island surrounded by the sea. In such an island, it is natural to ask a question: “Where is the most distant point from the sea?” The answer to this question for Honshu was found in 1996. The most distant point is located in former Usuda Town, Nagano Prefecture, whose distance from the sea is 114.86 km.
In this problem, you are asked to write a program which, given a map of an island, finds the most distant point from the sea in the island, and reports its distance from the sea. In order to simplify the problem, we only consider maps representable by convex polygons.
Input
The input consists of multiple datasets. Each dataset represents a map of an island, which is a convex polygon. The format of a dataset is as follows.
| n | ||
| x1 | y1 | |
| ⋮ | ||
| xn | yn |
Every input item in a dataset is a non-negative integer. Two input items in a line are separated by a space.
n in the first line is the number of vertices of the polygon, satisfying 3 ≤ n ≤ 100. Subsequent n lines are the x- and y-coordinates of the n vertices. Line segments (xi, yi)–(xi+1, yi+1) (1 ≤ i ≤ n − 1) and the line segment (xn, yn)–(x1, y1) form the border of the polygon in counterclockwise order. That is, these line segments see the inside of the polygon in the left of their directions. All coordinate values are between 0 and 10000, inclusive.
You can assume that the polygon is simple, that is, its border never crosses or touches itself. As stated above, the given polygon is always a convex one.
The last dataset is followed by a line containing a single zero.
Output
For each dataset in the input, one line containing the distance of the most distant point from the sea should be output. An output line should not contain extra characters such as spaces. The answer should not have an error greater than 0.00001 (10−5). You may output any number of digits after the decimal point, provided that the above accuracy condition is satisfied.
题目大意：逆时针给出一个凸多边形，求一个点，问这个点离凸多边形的边最远是多少。
思路：二分答案d，然后每条边向内平移d，看能否缩成一个点。
代码（POJ 32MS/UVA 22MS）：
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std; const int MAXN = ;
const double EPS = 1e-;
const double PI = acos(-1.0);//3.14159265358979323846 inline int sgn(double x) {
return (x > EPS) - (x < -EPS);
} struct Point {
double x, y, ag;
Point() {}
Point(double x, double y): x(x), y(y) {}
void read() {
scanf("%lf%lf", &x, &y);
}
bool operator == (const Point &rhs) const {
return sgn(x - rhs.x) == && sgn(y - rhs.y) == ;
}
bool operator < (const Point &rhs) const {
if(y != rhs.y) return y < rhs.y;
return x < rhs.x;
}
Point operator + (const Point &rhs) const {
return Point(x + rhs.x, y + rhs.y);
}
Point operator - (const Point &rhs) const {
return Point(x - rhs.x, y - rhs.y);
}
Point operator * (const int &b) const {
return Point(x * b, y * b);
}
Point operator / (const int &b) const {
return Point(x / b, y / b);
}
double length() const {
return sqrt(x * x + y * y);
}
Point unit() const {
return *this / length();
}
};
typedef Point Vector; double dist(const Point &a, const Point &b) {
return (a - b).length();
} double cross(const Point &a, const Point &b) {
return a.x * b.y - a.y * b.x;
}
//ret >= 0 means turn left
double cross(const Point &sp, const Point &ed, const Point &op) {
return sgn(cross(sp - op, ed - op));
} double area(const Point& a, const Point &b, const Point &c) {
return fabs(cross(a - c, b - c)) / ;
} struct Seg {
Point st, ed;
double ag;
Seg() {}
Seg(Point st, Point ed): st(st), ed(ed) {}
void read() {
st.read(); ed.read();
}
void makeAg() {
ag = atan2(ed.y - st.y, ed.x - st.x);
}
};
typedef Seg Line; void moveRight(Line &v, double r) {
double dx = v.ed.x - v.st.x, dy = v.ed.y - v.st.y;
dx = dx / dist(v.st, v.ed) * r;
dy = dy / dist(v.st, v.ed) * r;
v.st.x += dy; v.ed.x += dy;
v.st.y -= dx; v.ed.y -= dx;
} bool isOnSeg(const Seg &s, const Point &p) {
return (p == s.st || p == s.ed) ||
(((p.x - s.st.x) * (p.x - s.ed.x) < ||
(p.y - s.st.y) * (p.y - s.ed.y) < ) &&
sgn(cross(s.ed, p, s.st) == ));
} bool isIntersected(const Point &s1, const Point &e1, const Point &s2, const Point &e2) {
return (max(s1.x, e1.x) >= min(s2.x, e2.x)) &&
(max(s2.x, e2.x) >= min(s1.x, e1.x)) &&
(max(s1.y, e1.y) >= min(s2.y, e2.y)) &&
(max(s2.y, e2.y) >= min(s1.y, e1.y)) &&
(cross(s2, e1, s1) * cross(e1, e2, s1) >= ) &&
(cross(s1, e2, s2) * cross(e2, e1, s2) >= );
} bool isIntersected(const Seg &a, const Seg &b) {
return isIntersected(a.st, a.ed, b.st, b.ed);
} bool isParallel(const Seg &a, const Seg &b) {
return sgn(cross(a.ed - a.st, b.ed - b.st)) == ;
} //return Ax + By + C =0 's A, B, C
void Coefficient(const Line &L, double &A, double &B, double &C) {
A = L.ed.y - L.st.y;
B = L.st.x - L.ed.x;
C = L.ed.x * L.st.y - L.st.x * L.ed.y;
}
//point of intersection
Point operator * (const Line &a, const Line &b) {
double A1, B1, C1;
double A2, B2, C2;
Coefficient(a, A1, B1, C1);
Coefficient(b, A2, B2, C2);
Point I;
I.x = - (B2 * C1 - B1 * C2) / (A1 * B2 - A2 * B1);
I.y = (A2 * C1 - A1 * C2) / (A1 * B2 - A2 * B1);
return I;
} bool isEqual(const Line &a, const Line &b) {
double A1, B1, C1;
double A2, B2, C2;
Coefficient(a, A1, B1, C1);
Coefficient(b, A2, B2, C2);
return sgn(A1 * B2 - A2 * B1) == && sgn(A1 * C2 - A2 * C1) == && sgn(B1 * C2 - B2 * C1) == ;
} struct Poly {
int n;
Point p[MAXN];//p[n] = p[0]
void init(Point *pp, int nn) {
n = nn;
for(int i = ; i < n; ++i) p[i] = pp[i];
p[n] = p[];
}
double area() {
if(n < ) return ;
double s = p[].y * (p[n - ].x - p[].x);
for(int i = ; i < n; ++i)
s += p[i].y * (p[i - ].x - p[i + ].x);
return s / ;
}
}; void Graham_scan(Point *p, int n, int *stk, int &top) {//stk[0] = stk[top]
sort(p, p + n);
top = ;
stk[] = ; stk[] = ;
for(int i = ; i < n; ++i) {
while(top && cross(p[i], p[stk[top]], p[stk[top - ]]) >= ) --top;
stk[++top] = i;
}
int len = top;
stk[++top] = n - ;
for(int i = n - ; i >= ; --i) {
while(top != len && cross(p[i], p[stk[top]], p[stk[top - ]]) >= ) --top;
stk[++top] = i;
}
}
//use for half_planes_cross
bool cmpAg(const Line &a, const Line &b) {
if(sgn(a.ag - b.ag) == )
return sgn(cross(b.ed, a.st, b.st)) < ;
return a.ag < b.ag;
}
//clockwise
bool half_planes_cross(Line *v, int vn, Poly &res, Line *deq) {
int i, n;
sort(v, v + vn, cmpAg);
for(i = n = ; i < vn; ++i) {
if(sgn(v[i].ag - v[i-].ag) == ) continue;
v[n++] = v[i];
}
int head = , tail = ;
deq[] = v[], deq[] = v[];
for(i = ; i < n; ++i) {
if(isParallel(deq[tail - ], deq[tail]) || isParallel(deq[head], deq[head + ]))
return false;
while(head < tail && sgn(cross(v[i].ed, deq[tail - ] * deq[tail], v[i].st)) > )
--tail;
while(head < tail && sgn(cross(v[i].ed, deq[head] * deq[head + ], v[i].st)) > )
++head;
deq[++tail] = v[i];
}
while(head < tail && sgn(cross(deq[head].ed, deq[tail - ] * deq[tail], deq[head].st)) > )
--tail;
while(head < tail && sgn(cross(deq[tail].ed, deq[head] * deq[head + ], deq[tail].st)) > )
++head;
if(tail <= head + ) return false;
res.n = ;
for(i = head; i < tail; ++i)
res.p[res.n++] = deq[i] * deq[i + ];
res.p[res.n++] = deq[head] * deq[tail];
res.n = unique(res.p, res.p + res.n) - res.p;
res.p[res.n] = res.p[];
return true;
} /*******************************************************************************************/ Point p[MAXN];
Poly poly;
int stk[MAXN], top;
int n, T; Poly res; Line original[MAXN], newLine[MAXN], deq[MAXN]; bool check(double r) {
for(int i = ; i < n; ++i) newLine[i] = original[i];
for(int i = ; i < n; ++i) moveRight(newLine[i], r);
return half_planes_cross(newLine, n, res, deq);
} int main() {
while(scanf("%d", &n) != EOF && n) {
for(int i = ; i < n; ++i) p[i].read();
p[n] = p[];
for(int i = ; i < n; ++i) original[i] = Line(p[i + ], p[i]);
for(int i = ; i < n; ++i) original[i].makeAg();
double l = , r = 1e9;
for(int i = ; i < ; ++i) {
double mid = (l + r) / ;
if(check(mid)) l = mid;
else r = mid;
}
printf("%.10f\n", l);
}
}