鏈接：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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一開始看錯題意，WA了好多次，要注意與當前線段相鄰接的線段不判斷

主要就是第一個線段，要跳過與下一條線段的相交性，以及最后一條線段的相交性，其他線段只需要向下跳過一個線段判斷相交即可

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  1 #include <stdio.h>
  2 #include <string.h>
  3 #include <stdlib.h>
  4 #include <iostream>
  5 #include <algorithm>
  6 #include <math.h>
  7 
  8 #define MAXX 1005
  9 #define eps 1e-8
 10 using namespace std;
 11 
 12 typedef struct
 13 {
 14     double x;
 15     double y;
 16 }point;
 17 
 18 typedef struct
 19 {
 20     point st;
 21     point ed;
 22 }line;
 23 
 24 point p[MAXX];
 25 line li[MAXX];
 26 
 27 double crossProduct(point a,point b,point c)
 28 {
 29     return (c.x-a.x)*(b.y-a.y)-(c.y-a.y)*(b.x-a.x);
 30 }
 31 
 32 double dist(point a,point b)
 33 {
 34     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
 35 }
 36 
 37 bool  xy(double x,double y){ return x < y - eps; }
 38 bool  dy(double x,double y){ return x > y + eps; }
 39 bool xyd(double x,double y){ return x < y + eps; }
 40 bool dyd(double x,double y){ return x > y - eps; }
 41 bool  dd(double x,double y){ return fabs(x-y)<eps; }
 42 
 43 bool onSegment(point a,point b,point c)
 44 {
 45     double maxx=max(a.x,b.x);
 46     double maxy=max(a.y,b.y);
 47     double minx=min(a.x,b.x);
 48     double miny=min(a.y,b.y);
 49 
 50     if(dd(crossProduct(a,b,c),0.0)&&xyd(c.x,maxx)&&dyd(c.x,minx)
 51        &&xyd(c.y,maxy)&&dyd(c.y,miny))
 52         return true;
 53     return false;
 54 }
 55 
 56 bool segIntersect(point p1,point p2,point p3,point p4)
 57 {
 58     double d1=crossProduct(p3,p4,p1);
 59     double d2=crossProduct(p3,p4,p2);
 60     double d3=crossProduct(p1,p2,p3);
 61     double d4=crossProduct(p1,p2,p4);
 62 
 63     if(xy(d1*d2,0.0)&&xy(d3*d4,0.0))
 64         return true;
 65     if(dd(d1,0.0)&&onSegment(p3,p4,p1))
 66         return true;
 67     if(dd(d2,0.0)&&onSegment(p3,p4,p2))
 68         return true;
 69     if(dd(d3,0.0)&&onSegment(p1,p2,p3))
 70         return true;
 71     if(dd(d4,0.0)&&onSegment(p1,p2,p4))
 72         return true;
 73     return false;
 74 }
 75 
 76 double Area(int n)
 77 {
 78     double ans=0.0;
 79 
 80     for(int i=2; i<n; i++)
 81     {
 82         ans+=crossProduct(p[0],p[i-1],p[i]);
 83     }
 84     return fabs(ans)/2.0;
 85 }
 86 
 87 int main()
 88 {
 89     int n,m,i,j;
 90     double x,y;
 91     int cas=1;
 92     while(scanf("%d",&n)!=EOF&&n)
 93     {
 94         for(i=0; i<n; i++)
 95         {
 96             scanf("%lf%lf",&p[i].x,&p[i].y);
 97         }
 98 
 99         for(i=0; i<n-1; i++)
100         {
101             li[i].st.x=p[i].x;
102             li[i].st.y=p[i].y;
103             li[i].ed.x=p[i+1].x;
104             li[i].ed.y=p[i+1].y;
105         }
106         li[n-1].st.x=p[n-1].x;
107         li[n-1].st.y=p[n-1].y;
108         li[n-1].ed.x=p[0].x;
109         li[n-1].ed.y=p[0].y;
110         bool flag=false;
111         for(i=0; i<n; i++)
112         {
113             for(j=i+2; j<n; j++)
114             {
115                     if(i == 0 && j == n-1)continue;
116                     /*if((li[i].st.x == li[j].st.x && li[i].st.y == li[j].st.y)
117                        || li[i].st.x == li[j].ed.x && li[i].st.y == li[j].ed.y
118                        || li[i].ed.x == li[j].st.x && li[i].ed.y == li[j].st.y
119                        || li[i].ed.x == li[j].ed.x && li[i].ed.y == li[j].ed.y)
120                         continue;*/
121                     if(segIntersect(li[i].st,li[i].ed,li[j].st,li[j].ed))
122                     {
123                         flag=true;
124                         break;
125                     }
126             }
127         }
128 
129         if(flag || n<3)
130         {
131             printf("Figure %d: Impossible\n",cas++);
132         }
133         else
134         {
135             double ans=Area(n);
136             printf("Figure %d: %.2lf\n",cas++,ans);
137         }
138         printf("\n");
139     }
140     return 0;
141 }

View Code

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轉載于:https://www.cnblogs.com/ccccnzb/p/3903291.html

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