HDU 2056 Rectangles (求相交矩形的面積)

Rectangles

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 11631 Accepted Submission(s): 3716Problem DescriptionGiven two rectangles and the coordinates of two points on the diagonals of each rectangle,you have to calculate the area of the intersected part of two rectangles. its sides are parallel to OX and OY .

InputInput The first line of input is 8 positive numbers which indicate the coordinates of four points that must be on each diagonal.The 8 numbers are x1,y1,x2,y2,x3,y3,x4,y4.That means the two points on the first rectangle are(x1,y1),(x2,y2);the
other two points on the second rectangle are (x3,y3),(x4,y4).

OutputOutput For each case output the area of their intersected part in a single line.accurate up to 2 decimal places.

Sample Input

1.00 1.00 3.00 3.00 2.00 2.00 4.00 4.005.00 5.00 13.00 13.00 4.00 4.00 12.50 12.50

Sample Output

1.0056.25

題目大意  求相交矩形的面積

方法一:   矩形左上方 的座標和矩形右下角的座標

import java.io.*;import java.text.DecimalFormat;import java.util.*;public class Main {public static void main(String[] args) {Scanner sc=new Scanner(new BufferedInputStream(System.in));while(sc.hasNextDouble()){double s=0;double res[]=new double[4];double res1[]=new double[4];double res2[]=new double[4];double point1[]=new double[4];double point2[]=new double[4];for(int i=0;i<point1.length;i++){point1[i]=sc.nextDouble();}for(int i=0;i<point2.length;i++){point2[i]=sc.nextDouble();}//  第一個矩形左上方的座標res1[0]=Math.min(point1[0], point1[2]);//X1res1[1]=Math.max(point1[1], point1[3]);//Y1//第一個矩形右下角的座標res1[2]=Math.max(point1[0], point1[2]);//x2res1[3]=Math.min(point1[1], point1[3]);//y2//  第二個矩形左上方的座標res2[0]=Math.min(point2[0], point2[2]);//x1res2[1]=Math.max(point2[1], point2[3]);//y1// 第二個矩形右下角的座標res2[2]=Math.max(point2[0], point2[2]);//x2res2[3]=Math.min(point2[1], point2[3]);//y2// 如果矩形重合和相離  s=0  if(res2[0]>=res1[2]||res2[2]<=res1[0]||res2[3]>=res1[1]||res2[1]<=res1[3]){s=0;}else{//否者矩形相交的 座標res[0]=Math.max(res1[0], res2[0]);res[1]=Math.min(res1[1],res2[1]);res[2]=Math.min(res1[2],res2[2]);res[3]=Math.max(res1[3],res2[3]);s=(res[2]-res[0])*(res[1]-res[3]);}DecimalFormat fo=new DecimalFormat("0.00");System.out.println(fo.format(s));}}}

 

方法二   求矩形 左下角的座標和右上方的座標

import java.io.*;import java.text.DecimalFormat;import java.util.*;public class Main {public static void main(String[] args) {Scanner sc=new Scanner(new BufferedInputStream(System.in));while(sc.hasNextDouble()){double s=0;double x1=sc.nextDouble();double y1=sc.nextDouble();double x2=sc.nextDouble();double y2=sc.nextDouble();double x3=sc.nextDouble();double y3=sc.nextDouble();double x4=sc.nextDouble();double y4=sc.nextDouble();//第一個矩形左下角的座標double xx1=Math.min(x1,x2);//X1double yy1=Math.min(y1,y2);//Y1//第一個矩形右上方的座標double xx2=Math.max(x1,x2);//x2double yy2=Math.max(y1,y2);//y2//第二個矩形的左下角的座標double xx3=Math.min(x3,x4);//X1double yy3=Math.min(y3,y4);//Y1//第二個矩形右上方的座標double xx4=Math.max(x3,x4);//x2double yy4=Math.max(y3,y4);//y2// 如果矩形重合和相離  s=0  if(xx3>=xx2||xx4<=xx1||yy3>=yy2||yy1>=yy4){s=0;}else{//否者矩形相交的 座標double a=Math.max(xx1,xx3);double b=Math.min(xx2,xx4);double c=Math.max(yy1,yy3);double d=Math.min(yy2,yy4);s=(b-a)*(d-c);}DecimalFormat fo=new DecimalFormat("0.00");System.out.println(fo.format(s));}}}