#include<stdio.h>

    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,y1,x2,y2,x3,y3,x4,y4;
    double s;
    while(scanf("%lf %lf %lf %lf %lf %lf %lf %lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4)!=EOF)
    {
    if((min(x2,x4)-max(x1,x3)>=0)&& (min(y2,y4)-max(y1,y3)>=0))
        {
        s=(min(x2,x4)-max(x1,x3))* (min(y2,y4)-max(y1,y3));
        printf("%.2lf\n",s);
        }
    else
    printf("0.00\n");
    }
    return 0;
}
数据测试是对的 但是提交wa 求助哪里错了 射射!!

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




这个是题目！

“Given two rectangles and the coordinates of two points on the diagonals of each rectangle”

这里只说了是对角线上的点，但没说是哪条对角线。所以不要假设x1<x2这样的条件。

#include<stdio.h>
int main()
{
    double x1, y1, x2, y2, x3, y3, x4, y4, x, y;
    while(scanf("%lf %lf %lf %lf %lf %lf %lf %lf", &x1, &y1, &x2, &y2, &x3, &y3, &x4, &y4) != EOF)
    {
        if(x1 > x2){ x = x1; x1 = x2; x2 = x;}
        if(y1 > y2){ y = y1; y1 = y2; y2 = y;}
        if(x3 > x4){ x = x3; x3 = x4; x4 = x;}
        if(y3 > y4){ y = y3; y3 = y4; y4 = y;}
        x = (x4 > x2) ? x2 : x4;
        x -= (x3 > x1) ? x3 : x1;
        y = (y4 > y2) ? y2 : y4;
        y -= (y3 > y1) ? y3 : y1;
        if(x <= 0 || y <= 0)
            printf("0.00\n");
        else
            printf("%.2f\n", x * y);
    }
    return 0;
}

原来是这样 超感谢！

不客气，以后尽量少打错别字。