HDU2056JAVA

Posted 2020-07-13 折腾青春

Rectangles

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 20375    Accepted Submission(s): 6607

Problem Description

Given 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 .

 

 

Input

Input 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).

 

 

Output

Output 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 4.00

5.00 5.00 13.00 13.00 4.00 4.00 12.50 12.50

 

 

Sample Output

1.00

56.25

import java.util.Scanner;

 class Main{

public static void main(String[] args) {


Scanner cin=new Scanner(System.in);
while(cin.hasNext()){
double t;
double sum=0.0;
double x1=cin.nextDouble();
double y1=cin.nextDouble();
double x2=cin.nextDouble();
double y2=cin.nextDouble();
double x3=cin.nextDouble();
double y3=cin.nextDouble();
double x4=cin.nextDouble();
double y4=cin.nextDouble();
if(x1>x2){t=x1;x1=x2;x2=t;}
if(y1>y2){t=y1;y1=y2;y2=t;}
if(x3>x4){t=x3;x3=x4;x4=t;}
if(y3>y4){t=y3;y3=y4;y4=t;}
x1=max(x1,x3);
x2=min(x2,x4);
y1=max(y1,y3);
y2=min(y2,y4);
if(x1>x2||y1>y2){
System.out.println("0.00");
}
else{
sum=(x2-x1)*(y2-y1);
System.out.printf("%.2f",sum);
System.out.println();
}

}
}
private static double min(double x2, double x4) {
if(x2<x4)
return x2;
else
return x4;
}
private static double max(double x1, double x3) {
if(x1>x3)
return x1;
else
return x3;
}
}

这个地方要注意的是的判断一下图形是否相交，如果不相交，还是要输出按题目中的输出要求：0.00；不然就会出错，这个题目只有相交和分离两种状态；