zoj 1010 （线段相交判断+多边形求面积）

链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=10

Area

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Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge

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Jerry, a middle school student, addicts himself to mathematical research. Maybe the problems he has thought are really too easy to an expert. But as an amateur, especially as a 15-year-old boy, he had done very well. He is so rolling in thinking the mathematical problem that he is easily to try to solve every problem he met in a mathematical way. One day, he found a piece of paper on the desk. His younger sister, Mary, a four-year-old girl, had drawn some lines. But those lines formed a special kind of concave polygon by accident as Fig. 1 shows.

Fig. 1 The lines his sister had drawn

"Great!" he thought, "The polygon seems so regular. I had just learned how to calculate the area of triangle, rectangle and circle. I'm sure I can find out how to calculate the area of this figure." And so he did. First of all, he marked the vertexes in the polygon with their coordinates as Fig. 2 shows. And then he found the result--0.75 effortless.

Fig.2 The polygon with the coordinates of vertexes

Of course, he was not satisfied with the solution of such an easy problem. "Mmm, if there's a random polygon on the paper, then how can I calculate the area?" he asked himself. Till then, he hadn't found out the general rules on calculating the area of a random polygon. He clearly knew that the answer to this question is out of his competence. So he asked you, an erudite expert, to offer him help. The kind behavior would be highly appreciated by him.

Input

The input data consists of several figures. The first line of the input for each figure contains a single integer n, the number of vertexes in the figure. (0 <= n <= 1000).

In the following n lines, each contain a pair of real numbers, which describes the coordinates of the vertexes, (xi, yi). The figure in each test case starts from the first vertex to the second one, then from the second to the third, ���� and so on. At last, it closes from the nth vertex to the first one.

The input ends with an empty figure (n = 0). And this figure not be processed.

Output

As shown below, the output of each figure should contain the figure number and a colon followed by the area of the figure or the string "Impossible".

If the figure is a polygon, compute its area (accurate to two fractional digits). According to the input vertexes, if they cannot form a polygon (that is, one line intersects with another which shouldn't be adjoined with it, for example, in a figure with four lines, the first line intersects with the third one), just display "Impossible", indicating the figure can't be a polygon. If the amount of the vertexes is not enough to form a closed polygon, the output message should be "Impossible" either.

Print a blank line between each test cases.

Sample Input

5
0 0
0 1
0.5 0.5
1 1
1 0
4
0 0
0 1
1 0
1 1
0

Output for the Sample Input

Figure 1: 0.75

Figure 2: Impossible

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一开始看错题意，WA了好多次，要注意与当前线段相邻接的线段不判断

主要就是第一个线段，要跳过与下一条线段的相交性，以及最后一条线段的相交性，其他线段只需要向下跳过一个线段判断相交即可

 #include <stdio.h>

 #include <string.h>

 #include <stdlib.h>

 #include <iostream>

 #include <algorithm>

 #include <math.h>

 #define MAXX 1005

 #define eps 1e-8

 using namespace std;

 typedef struct

 {

     double x;

     double y;

 }point;

 typedef struct

 {

     point st;

     point ed;

 }line;

 point p[MAXX];

 line li[MAXX];

 double crossProduct(point a,point b,point c)

 {

     return (c.x-a.x)*(b.y-a.y)-(c.y-a.y)*(b.x-a.x);

 }

 double dist(point a,point b)

 {

     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));

 }

 bool  xy(double x,double y){ return x < y - eps; }

 bool  dy(double x,double y){ return x > y + eps; }

 bool xyd(double x,double y){ return x < y + eps; }

 bool dyd(double x,double y){ return x > y - eps; }

 bool  dd(double x,double y){ return fabs(x-y)<eps; }

 bool onSegment(point a,point b,point c)

 {

     double maxx=max(a.x,b.x);

     double maxy=max(a.y,b.y);

     double minx=min(a.x,b.x);

     double miny=min(a.y,b.y);

     if(dd(crossProduct(a,b,c),0.0)&&xyd(c.x,maxx)&&dyd(c.x,minx)

        &&xyd(c.y,maxy)&&dyd(c.y,miny))

         return true;

     return false;

 }

 bool segIntersect(point p1,point p2,point p3,point p4)

 {

     double d1=crossProduct(p3,p4,p1);

     double d2=crossProduct(p3,p4,p2);

     double d3=crossProduct(p1,p2,p3);

     double d4=crossProduct(p1,p2,p4);

     if(xy(d1*d2,0.0)&&xy(d3*d4,0.0))

         return true;

     if(dd(d1,0.0)&&onSegment(p3,p4,p1))

         return true;

     if(dd(d2,0.0)&&onSegment(p3,p4,p2))

         return true;

     if(dd(d3,0.0)&&onSegment(p1,p2,p3))

         return true;

     if(dd(d4,0.0)&&onSegment(p1,p2,p4))

         return true;

     return false;

 }

 double Area(int n)

 {

     double ans=0.0;

     for(int i=; i<n; i++)

     {

         ans+=crossProduct(p[],p[i-],p[i]);

     }

     return fabs(ans)/2.0;

 }

 int main()

 {

     int n,m,i,j;

     double x,y;

     int cas=;

     while(scanf("%d",&n)!=EOF&&n)

     {

         for(i=; i<n; i++)

         {

             scanf("%lf%lf",&p[i].x,&p[i].y);

         }

         for(i=; i<n-; i++)

         {

             li[i].st.x=p[i].x;

             li[i].st.y=p[i].y;

             li[i].ed.x=p[i+].x;

             li[i].ed.y=p[i+].y;

         }

         li[n-].st.x=p[n-].x;

         li[n-].st.y=p[n-].y;

         li[n-].ed.x=p[].x;

         li[n-].ed.y=p[].y;

         bool flag=false;

         for(i=; i<n; i++)

         {

             for(j=i+; j<n; j++)

             {

                     if(i ==  && j == n-)continue;

                     /*if((li[i].st.x == li[j].st.x && li[i].st.y == li[j].st.y)

                        || li[i].st.x == li[j].ed.x && li[i].st.y == li[j].ed.y

                        || li[i].ed.x == li[j].st.x && li[i].ed.y == li[j].st.y

                        || li[i].ed.x == li[j].ed.x && li[i].ed.y == li[j].ed.y)

                         continue;*/

                     if(segIntersect(li[i].st,li[i].ed,li[j].st,li[j].ed))

                     {

                         flag=true;

                         break;

                     }

             }

         }

         if(flag || n<)

         {

             printf("Figure %d: Impossible\n",cas++);

         }

         else

         {

             double ans=Area(n);

             printf("Figure %d: %.2lf\n",cas++,ans);

         }

         printf("\n");

     }

     return ;

 }

zoj 1010 （线段相交判断+多边形求面积）的相关教程结束。