Recovering 3-D structure from motion in noisy 2-D images is a problem addressed by many vision system researchers. By consistently tracking feature points of interest across multiple images using a methodology first described by Lucas-Kanade, a 3-D shape of the scene can be reconstructed using these features points using the factorization method developed by Tomasi-Kanade.
樣板 B 樹 ( B - tree )
規(guī)則 :
(1) 每個(gè)節(jié)點(diǎn)內(nèi)元素個(gè)數(shù)在 [MIN,2*MIN] 之間, 但根節(jié)點(diǎn)元素個(gè)數(shù)為 [1,2*MIN]
(2) 節(jié)點(diǎn)內(nèi)元素由小排到大, 元素不重複
(3) 每個(gè)節(jié)點(diǎn)內(nèi)的指標(biāo)個(gè)數(shù)為元素個(gè)數(shù)加一
(4) 第 i 個(gè)指標(biāo)所指向的子節(jié)點(diǎn)內(nèi)的所有元素值皆小於父節(jié)點(diǎn)的第 i 個(gè)元素
(5) B 樹內(nèi)的所有末端節(jié)點(diǎn)深度一樣
1. 下列說法正確的是 ( )
A. Java語言不區(qū)分大小寫
B. Java程序以類為基本單位
C. JVM為Java虛擬機(jī)JVM的英文縮寫
D. 運(yùn)行Java程序需要先安裝JDK
2. 下列說法中錯(cuò)誤的是 ( )
A. Java語言是編譯執(zhí)行的
B. Java中使用了多進(jìn)程技術(shù)
C. Java的單行注視以//開頭
D. Java語言具有很高的安全性
3. 下面不屬于Java語言特點(diǎn)的一項(xiàng)是( )
A. 安全性
B. 分布式
C. 移植性
D. 編譯執(zhí)行
4. 下列語句中,正確的項(xiàng)是 ( )
A . int $e,a,b=10
B. char c,d=’a’
C. float e=0.0d
D. double c=0.0f
給定兩個(gè)集合A、B,集合內(nèi)的任一元素x滿足1 ≤ x ≤ 109,并且每個(gè)集合的元素個(gè)數(shù)不大于105。我們希望求出A、B之間的關(guān)系。
任 務(wù) :給定兩個(gè)集合的描述,判斷它們滿足下列關(guān)系的哪一種:
A是B的一個(gè)真子集,輸出“A is a proper subset of B”
B是A的一個(gè)真子集,輸出“B is a proper subset of A”
A和B是同一個(gè)集合,輸出“A equals B”
A和B的交集為空,輸出“A and B are disjoint”
上述情況都不是,輸出“I m confused!”
We have a group of N items (represented by integers from 1 to N), and we know that there is some total order defined for these items. You may assume that no two elements will be equal (for all a, b: a<b or b<a). However, it is expensive to compare two items. Your task is to make a number of comparisons, and then output the sorted order. The cost of determining if a < b is given by the bth integer of element a of costs (space delimited), which is the same as the ath integer of element b. Naturally, you will be judged on the total cost of the comparisons you make before outputting the sorted order. If your order is incorrect, you will receive a 0. Otherwise, your score will be opt/cost, where opt is the best cost anyone has achieved and cost is the total cost of the comparisons you make (so your score for a test case will be between 0 and 1). Your score for the problem will simply be the sum of your scores for the individual test cases.