UVA 12386 smallest Polygon n points of arbitrary polygons for minimum perimeter science of violence

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Test instructions

Given n points, a polygon of n points (which can be concave polygons, but n points must be all on a polygon)

In all polygons that can consist of n points

The perimeter of the polygon with the smallest area-the minimum perimeter.

Ideas:

First we select a fixed point, then the next number of a permutation, there is (n-1)! permutations.

This sequence has a line between the two adjacent numbers.

Judge Polygon Legal: Any two line segments do not intersect. N^2

The rest is a simple update of the answer.

So the complexity is (n-1)! * N*n

#include <cstdio> #include <cstring> #include <algorithm> #include <set> #include <cmath>    using namespace Std;const int max_n = 2507;struct Point {double x, y;    Point (Double _x = 0, double _y = 0) {x = _x, y = _y;        } BOOL operator < (const point &AMP;RHS) Const {if (x! = rhs.x) return x < rhs.x;    Return y < Rhs.y; }};int n;int top;point st[100], P[100];typedef point point;double Cross (Point p0, point P1, point p2) {return (p1.x- p0.x) * (P2.Y-P0.Y)-(p2.x-p0.x) * (P1.Y-P0.Y);} Double Mult_cross (Point P1, point P2, point p0) {return (p1.x-p0.x) * (P2.Y-P0.Y)-(P1.Y-P0.Y) * (p2.x-p0.x);} BOOL Is_cross (Point-P1, point-P2, point-P3, point-P4) {if (Std::min (p1.x, p2.x) <= Std::max (p3.x, p4.x) && s Td::min (p3.x, p4.x) <= Std::max (p1.x, p2.x) &&std::min (p1.y, p2.y) <= Std::max (p3.y, P4.y) &AMP;&AMP;STD:: Min (p3.y, p4.y) <= Std::max (p1.y, P2.y) &&mult_cross (P1, P4, p3)* Mult_cross (P2, P4, p3) <= 0 &&mult_cross (P3, p2, p1) * Mult_cross (P4, P2, p1) <= 0) return True;return fals e;} Double dis (point p1, point p2) {return sqrt ((p1.x-p2.x) * (p1.x-p2.x) + (P1.Y-P2.Y) * (P1.Y-P2.Y));}    int B[100];bool Check () {b[n-1] = n-1;            for (int i = 0, i < n; ++i) {for (int j = 0; J < N; ++j) {if (i = = j) Continue;            int x = (i + 1)% n, y = (j + 1)% n;            if (i = = y | | j = = X | | x = = y) continue;        if (Is_cross (P[b[i]], p[b[x]], P[b[j]], p[b[y])) return false; }} return true; Inline Double Cross (const-point &a, const-point &b) {return a.x * b.y-a.y * b.x;} BOOL Equal (double A, double b) {return fabs (a) < 1e-6;}    int main () {int T;    scanf ("%d", &t);        while (t--> 0) {scanf ("%d", &n);            for (int i = 0; i < n; ++i) {int x, y;            scanf ("%d%d", &x, &y); p[i].x = 1. * x, p[I].Y = 1.        * y;            } if (n = = 3) {puts ("0.0000");        Continue        } for (int i = 0; i < n-1; ++i) b[i] = i;        Double minlength =-1, Minarea =-1, minarealength =-1;                Do {if (check ()) {Double area = 0, length = 0;                B[n] = b[0];                    for (int i = 0; i < n; i++) {length + = DIS (P[b[i]], p[b[i+1]);                Area + = Cross (P[b[i]], p[b[i+1]);              } area = fabs (area) * 0.5;               printf ("www%.4f%.4f\n", area, length); for (int i = 0; i < n; ++i) printf ("(%d%.f%.f)", B[i], p[b[i]].x, P[B[I]].Y);                Puts ("");                    if (minlength = =-1) {minlength = length;                    Minarea = area;                minarealength = length;                    } else {minlength = min (minlength, length); if (Minarea > Area | | (Equal (area, Minarea) && Length < minarealength) {Minarea = area;                    minarealength = length;        }}}//for (int i = 0; i < n; ++i) printf ("%d", b[i]);      } while (Next_permutation (b, B + n-1));        printf ("%.4f%.4f%.4f\n", minlength, Minarealength, Minarea);    printf ("%.4f\n", Fabs (Minarealength-minlength)); } return 0;}

UVA 12386 smallest Polygon n points of arbitrary polygons for minimum perimeter science of violence