Rectangles HDU - 2056（求两个矩形的面积交）

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对x与y分别排序就可以

#include <iostream>
#include <algorithm>

using namespace std;


struct node
{
    int id;
    double num;
};


bool cmp(node a,node b)
{
    if(a.num==b.num) 
        return a.id<b.id;
    return a.num<b.num;
}

int main()
{
    double x1,y1,x2,y2,x3,y3,x4,y4;
    while(scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4)!=EOF)
    {
        node t1[4];
        node t2[4];
        t1[0].num=x1;t2[0].num=y1;
        t1[1].num=x2;t2[1].num=y2;
        t1[2].num=x3;t2[2].num=y3;
        t1[3].num=x4;t2[3].num=y4;
        t1[0].id=t2[0].id=t1[1].id=t2[1].id=1;
        t1[2].id=t2[2].id=t1[3].id=t2[3].id=2;
        sort(t1,t1+4,cmp);
        sort(t2,t2+4,cmp);
        if(t1[0].id==t1[1].id||t2[0].id==t2[1].id)
        {
            printf("0.00\n");
            continue;
        }
        else
            printf("%.2lf\n",(t1[2].num-t1[1].num)*(t2[2].num-t2[1].num));
        
    }
    return 0;
}