Ultraviolet A Problem 922-rectangle by the ocean

// Ultraviolet (a) problem 922-rectangle by the ocean // verdict: accepted // submission date: 2011-11-08/ultraviolet (a) Run Time: 0.028 S // All Rights Reserved (c) 2011, Qiu. Metaphysis # Yeah dot net /// [problem solving method] // This topic describes how to calculate the area of a polygon by using a directed area. Use the exhaustive method to enumerate the coordinates in the lower left corner and upper right corner of the rectangle to determine whether the rectangle has at least three corners on the polygon. Pay attention to the sequence of coordinates selected during enumeration, you can select the appropriate option to automatically obtain the // result of the minimum Lexicographic Order. # Include <iostream> # include <cmath> using namespace STD; # define maxpoly 300 struct point {int X; int y ;}; struct rect {int left, lower; int right, upper ;}; struct polygon {int N; int top, left, right, bottom; point P [maxpoly] ;}; // whether the point (I, j) is on a polygon. Int on [maxpoly] [maxpoly]; // use the exhaustive method to enumerate the coordinates in the lower left corner and upper right corner of a rectangle one by one. Void rectangle (polygon & contour, double area) {double mindiff = Area, currentdiff; rect TMP; // initialization. For (INT I = 0; I <= (contour. right-contour. left); I ++) for (Int J = 0; j <= (contour. top-contour. bottom); j ++) on [I] [J] = 0; // mark a point on a polygon as a polygon. In addition, points in the external rectangle of a polygon are not on the polygon. Note // coordinate transformation. For (INT I = 0; I <contour. n; I ++) on [contour. P [I]. x-contour. left] [contour. P [I]. y-contour. bottom] = 1; // enumerate possible rectangles. Pay attention to the enumeration order. For (INT I = 0; I <= contour. right-contour. left; I ++) for (Int J = 0; j <= contour. top-contour. bottom; j ++) for (int m = I; m <= contour. right-contour. left; m ++) for (INT n = J; n <= contour. top-contour. bottom; n ++) {// determines whether the condition of at least three corners on the polygon is met. If (on [I] [J] + on [I] [N] + on [m] [J] + on [m] [N] <3) continue; // compare the difference value. Note that it is "as close as possible" and the absolute value should be taken. Currentdiff = FABS (m-I) * (N-j)-area); If (currentdiff <mindiff) {mindiff = currentdiff; TMP = (rect) {contour. left + I, contour. bottom + J, contour. left + M, contour. bottom + N };}// output result. Cout <(INT) (area) <". "<(INT) (area * 10.0) % 10) <" "; cout <TMP. left <"" <TMP. lower <"; cout <TMP. right <"" <TMP. upper <Endl ;}// calculate the area of the polygon based on the vertex coordinates of the polygon. Double calarea (polygon & contour) {double Total = 0.0; For (INT I = 0; I <contour. n; I ++) {Int J = (I + 1) % contour. n; Total + = (contour. P [I]. x * contour. P [J]. y-contour. P [J]. x * contour. P [I]. y);} return FABS (total/2.0);} int main (int ac, char * AV []) {INT cases; polygon contour; CIN> cases; while (cases --) {CIN> contour. n; For (INT I = 0; I <contour. n; I ++) CIN> contour. P [I]. x> contour. P [I]. y; // find multiple Edge Shape external rectangular boundary. Contour. top = contour. P [0]. y; contour. left = contour. P [0]. x; contour. bottom = contour. P [0]. y; contour. right = contour. P [0]. x; For (INT I = 1; I <contour. n; I ++) {contour. top = max (contour. top, contour. P [I]. y); contour. left = min (contour. left, contour. P [I]. x); contour. bottom = min (contour. bottom, contour. P [I]. y); contour. right = max (contour. right, contour. P [I]. x);} rectangle (contour, calarea (contour);} return 0 ;}