6个回答

Book1346 每比对一次，都能得到一段新的不重叠的A时间段（刚没考虑太多，也可能出现两端不重叠时间段，所以可以采用递归算法进行比对）

1、按照你的描述不重叠时间段，最终的输出是：b={1,2}
2、还是要做计算输出其他呢？

1)我们先约定用`a[B,E]`方式表示时间段结构a，起始时间是a.B，结束时间是a.E。
2)讨论取`a[B,E]``b1[B,E]`不重叠部分：
2.a)当 `(a.E<=b1.B)||(b1.E<=a.B)` 时，无重叠，保留原先的a
2.b)否则有重叠，去掉原先的a；
`(a.B<b1.B)&&(b1.B<=a.E)` 时，留下左边非重复段 `a1[a.B,b1.B]`
`(a.B<=b1.E)&&(b1.E<a.E)` 时，留下右边非重复段 `a2[b1.E,a.E]`
3)对上面留下的每个a，都按找2)的逻辑和下个b2去重叠。
4)重复直到没有留下任何a或所有的b都用完了。

Tiger_Zhao 回复站在星空下: 编程需要用两个集合，处理前、处理后。拿到一个b后需要对处理前集合取出的每个a进行处理对照，无论留下几段都放到处理后集合中。拿下个b时，交换两个集合。

Tiger_Zhao 回复站在星空下: 没错，`2.b)`的3句话是并列的，可以同时有`a1``a2`留下啊。

Tiger_Zhao 缩进没了！`2.b)`包括后面的两行。自己想象缩进吧。

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Problem Description To see his girl friend, sweet potato has to go over thousands of mountains. What make things worse, many bitter potatoes lives in these mountains. They hate sweet potato because they don't have girl friends. In the world of potatoes, friction force does not exist. So the way potatoes travel is very simple: they start with an initial speed, rolling forward and waiting to get to the destination. Bitter potatoes lived in different places. When sweet potato rolls passing their home, they begin to chase him (surely by rolling with an initial speed). If sweet potato is caught by a bitter potato, his date with girl friend has to be canceled. Now sweet potato wants to know the minimum initial speed necessary to see his girl friend. Input First line is an integer T (T ≤ 50), the number of test cases. At the beginning of each case is three integers, N, M and w, indicate the number of peaks in the mountains, the number of bitter potatoes and the weight of sweet potato separately. 2 ≤ N ≤ 1000, 0 ≤ M ≤ 1000, 0 < w < 100000. The next N lines each contain a pair of integers. Each pair of integers xi, hi describe a peak. xi is the horizontal distant between sweet potato's home and the peak. hi is the height of the peak. All xi are different. 0 = x1 < x2 < … < xn ≤ 100000000, -100000000 ≤ hi ≤ 100000000. Between adjacent peaks is a smooth slope. The bitter potatoes are on these slopes. The following M lines each contain 3 integers. Each triple of integers pi, vi, mi describe a bitter potato. pi is the horizontal distant between his home and sweet potato’s home. vi is his initial speed. mi is his weight. 0 < pi < xn, 0 ≤ vi ≤ 100000000, 0 < mi < 100000 The gravitational constant in potatoes' world is 20. Sweet potato's home is at point (x1, h1). His girl friend lives at point (xn, hn). Output For each case, you should output “Case k: ” first. Following a number, the lower bound of sweet potato's initial speed rounded to two decimal places. Sample Input 1 6 2 100 0 0 2 5 3 2 4 1 5 3 8 -2 2 15 100 5 11 100 Sample Output Case 1: 20.62
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Problem Description There is a big round cake on the ground. A small ant plans to steal a small piece of cake. He starts from a certain point, reaches the cake, and then carry the piece back home. He does not want to be detected, so he is going to design a shortest path to achieve his goal. The big cake can be considered as a circle on a 2D plane. The ant’s home can be considered as a rectangle. The ant can walk through the cake. Please find out the shortest path for the poor ant. Input The input consists of several test cases. The first line of each test case contains x,y, representing the coordinate of the starting point. The second line contains x, y, r. The center of the cake is point (x,y) and the radius of the cake is r. The third line contains x1,y1,x2,y2, representing the coordinates of two opposite vertices of the rectangle --- the ant's home. All numbers in the input are real numbers range from -10000 to 10000. It is guaranteed that the cake and the ant's home don't overlap or contact, and the ant's starting point also is not inside the cake or his home, and doesn't contact with the cake or his home. If the ant touches any part of home, then he is at home. Input ends with a line of 0 0. There may be a blank line between two test cases. Output For each test case, print the shortest distance to achieve his goal. Please round the result to 2 digits after decimal point. Sample Input 1 1 -1 1 1 0 -1 1 0 0 2 -1 1 1 0 -1 1 0 0 0 Sample Output 1.75 2.00

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V代表点，E代表边。因为要找的是时间复杂度为O(n(V+E))的算法，所以传统的kruskal和prims并不符合。求一个算法思路可以满足O(n(V+E))的，不需要具体代码，非常感谢！！！！！
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C语言 算法：打分问题！！！
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