ZOJ 1010 Area

ZOJ Problem Set - 1010

Area

Time Limit: 2 Seconds

Memory Limit: 65536 KB

Special Judge

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.


1010 Area" TITLE="ZOJ 1010 Area" />

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.


1010 Area" TITLE="ZOJ 1010 Area" />

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

Source: Asia 2001, Shanghai (Mainland China)

源码：



1010 Area" TITLE="ZOJ 1010 Area" />



1010 Area" TITLE="ZOJ 1010 Area" />



1010 Area" TITLE="ZOJ 1010 Area" />

解题报告：

题目大意：对于任意给出的点，计算出他们的所围成的多边形面积，同时也要判断该多边形是否满足

条件。

算法思想：本题是数学题，较为复杂，单独的计算多边形的面积比较简单，对于三角形的面积

计算，已知三点a,b,c，ab叉乘ac的绝对值即为面积的一半，可以推导出多边形的面积公式。

多边形的顶点p0,p1,p2,p3,p4....,S=0.5*(p0*p1+p1*p2+p3*p4....+pn*p0).

其次，还要判断该多边形是否满足，判断是否任意两边会相交，任意两边的顶点会重合这些都不满足要求

由题意，该多边形的构造是在给出的点的顺序连线而成。所以任意两边我们可以直接用下标来表示:s,t,表示的是边（s,s+1）和边（t,t+1）

判断（s,s+1）和（t,t+1）是否相交，判断t点和t+1点是否在边（s,s+1）两侧，可以用叉乘判断，X叉乘Y和Y叉乘X符号相反

可以利用这个性质判断，只要看sXt sX(t+1) 符号是否相反。

判断（s,s+1）和（t,t+1）是否有重点时，有两个标准可以判断，对于s，t，t+1（1）如果按上述计算的叉乘=0，说明三点在一条线上，（2）如果他们的点乘=0，说明垂直，如果同时满足以上条件即可说明有重点。

以上的判断，都要做4次，分别以s，s+1，t，t+1,做参考。

上面的工作做完后，就要对该多边形的所有两两不同的边进行判断，需要注意的是，x[0],y[0]存放了点n的坐标，所以最后一步不用判断。即边（0,1）和边（n-1，n）肯定相交。