hdoj 2056 Rectangles（矩形） 【几何】

Rectangles

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 20290    Accepted Submission(s): 6574

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 4.005.00 5.00 13.00 13.00 4.00 4.00 12.50 12.50

 

Sample Output

1.0056.25
这道题开始想的有点乱！！看了详解，！！题意：给出两个矩形（任一）对角线的坐标，求两矩形重叠部分的面积。主要找坐标！#include <cstdio>#include <algorithm>#include <iostream>#define PI 3.1415926using namespace std;double max(double a,double b){    return a>b?a:b;}double min(double a,double b){    return a<b?a:b;}int main(){    double x1,x2,x3,x4,y1,y2,y3,y4;    while(~scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4))    {        double t;        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);        y1=max(y1,y3);        x2=min(x2,x4);        y2=min(y2,y4);//        if(y2<y1||x2<x1)//            printf("0.00\n");//        else//            printf("%.2lf\n",(y2-y1)*(x2-x1));        if(y2>=y1&&x2>=x1)            printf("%.2lf\n",(y2-y1)*(x2-x1));        else            printf("0.00\n");    }    return 0;}