HDU 3103 Shoring Up the Levees(计算几何 求面积)

标签：计算几何   面积   

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3103

The tiny country of Waterlogged is protected by a series of levees that form a quadrilateral as shown below:




The quadrilateral is defined by four vertices. The levees partition the country into four quadrants. Each quadrant is identified by a pair of vertices representing the outside edge of that quadrant. For example, Quadrant 1 shown below is defined by the points (x1, y1) and (x2, y2) .




It happens very often that the country of Waterlogged becomes flooded, and the levees need to be reinforced, but their country is poor and they have limited resources. They would like to be able to reinforce those levees that encompass the largest area first, then the next largest second, then the next largest third, and the smallest area fourth.

Help Waterlogged identify which quadrants are the largest, and the length of the levees around them.

1 2 1 5 5 2 2 0
3.5 2.2 4.8 -9.6 -1.2 -4.4 -8.9 12.4
0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

5.100 11.459 3.400 9.045 0.900 6.659 0.600 4.876
44.548 38.972 21.982 25.997 20.342 38.374 10.038 19.043

题意：

给出四个点，连接对角线后，分为四个象限，按照面积大小依次输出，如果面积相同则按照周长大小输出（注意：比较面积是否相同是比较保留了三位后是否相同）；

代码如下：

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
const double eps = 1e-5;
const double PI = acos(-1.0);

struct point
{
    double x, y;
};
struct gao
{
    double mz,zc;
};
struct gao gg[10];

bool cmp(gao a,gao b)
{
    if(a.mz!=b.mz)
        return a.mz>b.mz;
    return a.zc>b.zc;
}
double xmult(double x1,double y1,double x2,double y2,double x0,double y0)
{
    return (x1-x0)*(y2-y0)-(x2-x0)*(y1-y0);
}

//判两点在线段同侧,点在线段上返回0
int same_side(point p1,point p2,point l1,point l2)
{
    return xmult(l1.x,l1.y,p1.x,p1.y,l2.x,l2.y)*xmult(l1.x,l1.y,p2.x,p2.y,l2.x,l2.y)>0;
}

//两点距离
double dis(point a,point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
//两线段的交点
point intersection(point u1,point u2,point v1,point v2)
{
    point ret=u1;
    double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))
             /((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));
    ret.x+=(u2.x-u1.x)*t;
    ret.y+=(u2.y-u1.y)*t;
    return ret;
}

//三点面积
double aera(point a,point b,point c)
{
    double aa,bb,cc,q;
    aa=dis(c,b);
    bb=dis(a,c);
    cc=dis(b,a);
    q=(aa+bb+cc)/2;
    double h=sqrt(q*(q-aa)*(q-bb)*(q-cc));
    h=(int)(h*1000+0.5);
    return h*0.001;
}

//三点周长
double get_zc(point a,point b,point c)
{
    double aa,bb,cc,q;
    aa=dis(c,b);
    bb=dis(a,c);
    cc=dis(b,a);
    q=(aa+bb+cc);
    return q;
}


int main()
{
    int i;
    double x1,y1,x2,y2,x3,y3,x4,y4;
    point a,b,c,d,e;
    while(scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4)!=EOF)
    {
        if(x1==0 && y1==0 && x2==0 && y2==0 && x3==0 && y3==0 && x4==0 && y4==0)
            break;
        a.x=x1;
        a.y=y1;
        b.x=x2;
        b.y=y2;
        c.x=x3;
        c.y=y3;
        d.x=x4;
        d.y=y4;
        if(same_side(a, b, c,d)==0)
            e = intersection(d,c,a,b);
        else if(same_side(d, b, c, a)==0)
            e = intersection(d,b,c,a);
        else
            e = intersection(b,c,d,a);
        gg[0].mz=aera(a,b,e);
        gg[1].mz=aera(b,c,e);
        gg[2].mz=aera(c,d,e);
        gg[3].mz=aera(a,d,e);
        gg[0].zc=get_zc(a,b,e);
        gg[1].zc=get_zc(b,c,e);
        gg[2].zc=get_zc(c,d,e);
        gg[3].zc=get_zc(a,d,e);
        sort(gg,gg+4,cmp);
        for(i=0; i<3; i++)
            printf("%.3lf %.3lf ",gg[i].mz,gg[i].zc);
        printf("%.3lf %.3lf\n",gg[i].mz,gg[i].zc);
    }
    return 0;
}
/*
2 0 2 2 0 2 0 0
*/

HDU 3103 Shoring Up the Levees(计算几何 求面积)

标签：计算几何   面积   

原文地址：http://blog.csdn.net/u012860063/article/details/39831391