USACO section 3.4 Closed Fences(计算几何+叉积+二分)_「usaco3.4」closed fences-优快云博客

本文链接：https://blog.youkuaiyun.com/nealgavin/article/details/8690562

该博客讨论了如何检查一个由非交叉线段构成的封闭篱笆的有效性，并找出从特定点(x, y)可以看到的篱笆边。通过计算几何中的叉积和二分法，判断从给定点出发的光线是否能到达篱笆边且不与其他边相交。文章中提到了一道编程题目，涉及线段可见性的判断算法，并分享了作者在解决此问题时遇到的困难和解决方案。" 105588488,7446128,"AVL树的旋转操作详解：LL, RR, LR, RL

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Closed Fences

A closed fence in the plane is a set of non-crossing, connected line segments with N corners (3 < N < 200). The corners or vertices are each distinct and are listed in counter-clockwise order in an array {x_i, y_i}, i in (1..N).

Every pair of adjacent vertices defines a side of the fence. Thus {x_i y_i x_i+1 y_i+1} is a side of the fence for all i in (1..N). For our purposes, N+1 = 1, so that the first and last vertices making the fence closed.

Here is a typical closed fence and a point x,y:

                         * x3,y3
                 x5,y5  / \
    x,y *          *   /   \
                  / \ /     \
                 /   *       \
           x6,y6*   x4,y4     \
                |              \
                |               \
           x1,y1*----------------* x2,y2

Write a program which will do the following:

Test an ordered list of vertices {x_i,y_i}, i in (1..N) to see if the array is a valid fence.
Find the set of fence sides that a person (with no height) who is standing in the plane at position (x,y) can "see" when looking at the fence. The location x,y may fall anywhere not on the fence.

A fence side can be seen if there exists a ray that connects (x,y) and any point on the side, and the ray does not intersect any other side of the fence. A side that is parallel to the line of sight is not considered visible. In the figure, above the segments x3,y3-x4,y4; x5,y5-x6,y6; and x6-y6-x1,y1 are visible or partially visible from x,y.

PROGRAM NAME: fence4

INPUT FORMAT

Line 1:	N, the number of corners in the fence
Line 2:	Two space-separated integers, x and y, that are the location of the observer. Both integers will fit into 16 bits.
Line 3-N+2:	A pair of space-separated integers denoting the X,Y location of the corner. The pairs are given in counterclockwise order. Both integers are no larger than 1000 in magnitude.

NOTE: I have added anNew test case #12 for this task. Let me know if you think it's wrong. Rob Be sure to include USACO in your mail subject!

SAMPLE INPUT (file fence4.in)

OUTPUT FORMAT

If the sequence is not a valid fence, the output is a single line containing the word "NOFENCE".

Otherwise, the output is a listing of visible fence segments, one per line, shown as four space-separated integers that represent the two corners. Express the points in the segment by showing first the point that is earlier in the input, then the point that is later. Sort the segments for output by examining the last point and showing first those points that are earlier in the input. Use the same rule on the first of the two points in case of ties.

SAMPLE OUTPUT (file fence4.out)

思路：直接枚举每条边，判断这条边是否能看见，判断方法是：从人所在的位置e 分别与需要判段的边的两端连线，然后判断这两条线是否与其他的边同时非严格相交（可交与边界点），如果是，则这条边肯定被挡住了，否则在看是否有一条连线与其严格相交，有就将这条边砍成两段，继续判断，只要砍出一段不被挡住，这条边就不算被挡住，如果砍到很短的值还找不到，那就算被挡住了。

吐槽下：这道题卡了整一个多月，当初不咋会几何，而且，还得用二分判断特殊的情况，这题确实挺难，或许在大神眼中就是道水水吧。