最大效用频谱分配算法
关于认知无线电频谱分配方面的知识，希望对寻找资源的同学有帮助第37卷第19期何利,郑湘渝,刘振坤:基于图着色理论的最大效用频谱分配算法9542最大平均效用ISAA,但小于CSGC的时问开销。但是由图2可知,在同一图2分别给出了CSGC、ISAA和MUAA3种算法在用频谱数目下,MUAA算法的最大平均效用是最高的,并且随户取值为7和12时的最大平均效用。随着系统中频谱数的不着系统中频谱数目的増多,MUAA算法的最大平均效用高于淅增加,3种算法的最大平均效用基本呈现上升趋势,但是SAA算法,ISAA算法和(S(C算法的效用相同。MUAA算MUAA算法的最大平均效用増加趋势大于CSGC和ISAA算法以牺牲较少的时间开销获得了较大的系统效用,可见,法,这是因为随着频谱数目增加,MUAA算法的用户提高了MUAA算法是这3种算法中最优的。频谱的复用率。5结束语120本文研究了认知无线电网络屮基于图论着色模型的频谱H-ISAAMUAA分配算法,在分析CSGC、ISAA存在的问题基础上,提出兼CSGC顾系统效用和时间开销的MUAA算法,并对3种算法在系统效用和时间开销方面进行仿真比较。仿真结果表明,MUAA算法是有效的。下一步工作是将ISAA的并行计算方法应用2个到MUAA屮,在不增加时间开销的基础上,提高频谱分配的系统效用。参考文献[1] Shared Spectrum Company. Dynamic Spectrum Use[EB/OLI频谱数日(2007-04-25).http://www.sharedspectrum.com(an=7[2]Zheng Haitao, Peng Chunyi. Collaboration and Fairness inOpportunistic Spectrum Access[C]/Proc. of the 40th AnnuaMUAAInlernatiunal Conference on Communications. Seoul Kurea: Ieee3]廖楚林,陈劼,唐有喜,等.认知无限电中的并行频谱分配算法[J.电子与信息学报,2007,29(7:1608-1611[4 Wang Jiao, Huang Yuqing, Jiang Hong. Improved algorithm ofSpectrum Allocation Based on Graph Coloring Model in CognitiveRadio[clproc. of International Conference on Communicationsand Mubile Conputing. Washington D. C, USA: IEEE CoInputer频谱数目ociety,2009:353-357[S]彭振,赵知劲.基于混合蛙跣算法的认知无线电频谱分配[门.图2CsGC、ISAA和MUAA的最大平均效用计算机工程,2010,36(6):210-212.综上所述,由图1可知,MUAA算法的时间开销稍大」编辑陆燕菲(上接第92页信道下本地频谱感知和认知网络合作频谱感知进行研究和仿真,结果表明,合作频谱感知能明显改善 Rician衰洛信道的检测效果。下一步工作是研究其他衰落信道如 Nakagami信道下无线认知网络的合作频谱感知参考文献[]周贤伟.软件无线电M].北京:国防工业出版社,2008AWGN[2 Ghascmi A, Sousa E s. Collaborative Spcctrum Sensing forOpportunistic Access in Fading Environments[C]//Proc. of the Ist10IEEE SYmp. on Dynamic Spectrum Access Networks, Baltimore,103锴误警报概率USA: IEEE Press. 2005: 131-136图4不同用户数目时合作频谱感知的ROC曲线[3] Digham FF, Alouini M S, Simon M K On the Energy Detection of4绪束语Unknown Signals over Fading Channels[C]/Proc. of ICC"03在深度阴影衰落环境屮,认知用户需要合作频谱感知来Ottawa Canada: IEEE Press. 2003: 3575-3579检测到主用户的存在。 Rician信道模型适用于郊区或农村建4 Fctc b a.认知无线电技术lMJ.赵知劲,郑什链,尚俊娜,译模,以后将可能广泛应用,因此,对 Rician衰落信道的认知北京:科学出版社,2008网络合作频谱感知的研究具有重要意义。本文对 Rician衰落编辑金胡考C1994-2012ChinaAcademicJournalElcctronicPublishinghOusc.Allrightsrescrved.http://www.cnki.nct
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LeetCode cpp最新中文题解.pdf
LeetCode cpp最新中文题解.pdfLeetCode cpp最新中文题解.pdfLeetCode cpp最新中文题解.pdf目录3.4 Add binary615.1.5 Binary Tree Level Or-3.5 Longest Palindromic Substring. 62der traversal il3.6 Regular Expression Matching665.1.6 Binary Tree Zigzag3.7 Wildcard Matching67Level Order traversal. 963.8 Longest Common Prefix5.1.7 Recover Binary Search3. 9 Valid Number70Tree983.10 Integer to roman725. 1. 8 Same Tree3. 11 Roman to Integer735.1.9 Symmetric Tree1003.12 Count and Say745.1.10 Balanced Binary Tree.. 1023. 13 Anagrams755.1.11 Flatten Binary Tree to3. 14 Simplify Path76Linked List1033. 15 Length of Last Word775.1. 12 Populating Next RightPointers in each node ii 105第4章栈和队列7952二叉树的构建10641栈795.2.1 Construct Binary Tree4Valid Parentheses79from Preorder and In4.1.2 Longest valid Parenorder Traversa106theses805.2.2 Construct Binary Tree4.1.3 Largest Rectangle infrom Inorder and posHistogram82torder Traversal1074.1.4 Evaluate reverse pol-53二叉查找树108ish notation845.3. 1 Unique Binary Search4.2队列85Trees.1085.3.2 Unique Binary Search第5章树86Trees li.1105.1二叉树的遍历865.3.3 Validate Binary Search5.1.1 Binary Tree PreorderTreeTraversal865.3. 4 Convert Sorted array to5.1.2 Binary Tree InorderBinary Search Tree...112Traversal885.3.5 Convert Sorted List to5.1. 3 Binary Tree PostorderBinary Search Tree113Traversal9054二叉树的递归.1145. 1. 4 Binary Tree Level Or5.4.1 Minimum Depth of Bider traversal)2nary lree115目录5.4.2 Maximum Depth of Bi8.3.,2重新实现 next permunary Tree116tation1425.4.3 Path Sum11783.3递归.1435.4 4 Path Sum il118 8.4 Permutations II1445.4.5 Binary Tree Maximum8.4.1 next permutation... 144Path Suum11984.2重新实现 next permu5.4.6 Populating Next Righttation144Pointers in each node 120843递归1445.4.7 Sum Root to Leaf num8.5 Combinations146bers2185.1递归1468.5.2迭代147第6章排序1238.6 Letter Combinations of a phone6.1 Merge Sorted Array123umber1476.2 Merge Two Sorted Lists12486.1递归1486.3 Merge k Sorted Lists124862迭代96.4 Insertion Sort List125第9章广度优先搜索1506.5 Sort list1269.1 Word Ladder1506.6 First Missing Positive1279.2 Word Ladder il1546.7 Sort Colors1289.3 Surrounded regions162第7章查找94小结16413194.l适用场景1647.1 Search for a range131942思考的步骤7.2 Search Insert Position.13294.3代码模板1657. 3 Search a 2D Matrix133第10章深度优先搜索173第8章暴力枚举法13510.1 Palindrome Partitioning1738.1 Subsets13510.2 Unique Paths1768.1.1递归1350.2.1深搜1768.1.2迭代.1371022备忘录法.1768.2 Subsets il13810.23动规177821递归13810.24数学公式1788.2.2迭代.14110.3 Unique Paths Il1798. 3 Permutations14210.3.1备忘录法1798.3.1 next permutation14210.3.2动规.180目录10.4 N-Queens1813.4 Maximal rectangle21310.5 N-Queens II18413.5 Best Time to Buy and Sell Stock10.6 Restore ip addresses186.21410.7 Combination Sum18813.6 Interleaving String21510.8 Combination Sum Il18913.7 Scramble String21710.9 Generate Parentheses.19013. 8 Minimum Path Sum.22210.10 Sudoku solver19213.9 Edit Distance22410.11 Word Search.19313. 10 Decode Ways.22610.12小结19513. 11 Distinct sub22710.12.1适用场景19513. 12 Word Break22810.122思考的步骤19513 13 Word Break il2300.12.3代码模板197第14章图23210.12.4深搜与回溯法的区別.19714. 1 Clone Graph23210.12.5深搜与递归的区别..197第15章细节实现题235第11章分治法19915.1 Reverse Integer2351.1 Pow(x, n)19915.2 Palindrome Number.23611. 2 Sqrt(x)20015.3 Insert Interval237第12章贪心法20115.4 Merge Intervals23812.1 Jump game20115.5 Minimum Window Substring23912.2 Jump game II15.6 Multiply Strings24112. 3 Best Time to buy and Sell stock 20415.7 Substring with Concatenation12. 4 Best Time to buy and sell stock l205of all words24412. 5 Longest Substring Without re15.8 Pascal,s Triangle245peating Characters20615.9 Pascals Triangle Il24612.6 Container with Most Water.. 207 15.10 Spiral Matrix24715.11 Spiral matrix II248第13章动态规划20915.12 ZigZag Conversion25013. 1 Triangle20915.13 Divide Two Integers25113.2 Maximum Subarray15. 14 Text Justification25313.3 Palindrome Partitioning II1215.15 Max Points on a line255目录第1章编程技巧在判断两个浮点数a和b是否相等时,不要用a==b,应该判断二者之差的绝对值fabs(a-b)是否小于某个阈值,例如1e-9。判断一个整数是否是为奇数,用x%2!=0,不要用x%2=1,因为ⅹ可能是负用char的值作为数组下标(例如,统计字符串中每个字符岀现的次数),要考虑到char可能是负数。有的人考虑到了,先强制转型为 unsigned int再用作下标,这仍然是错的。正确的做法是,先强制转型为 unsigned char,再用作下标。这涉及C十整型提升的规则,就不详述了。以下是关于STL使用技巧的,很多条款来自《 Effective STL》这本书。vector和 string优先于动态分配的数组首先,在性能上,由于 vector能够保证连续内存,因此一旦分配了后,它的性能跟原始数组相当;其次,如果用new,意味着你要确保后面进行孓 delete,一旦忘记了,就会出现BUG,且这样需要都写一行 delete,代码不够短再次,声明多维数组的话,只能一个一个new,例如int** ary = new int*[row_num];for(int i=0: i< row num; ++1)ary [i] new int [col_num]用 vector的话一行代码搞定vectorary(row_num, vector(col_num, 0))使用 reserve来避免不必要的重新分配第2章线性表这类题目考察线性表的操作,例如,数组,单链表,双向链表等。2数组2.1.1 Remove Duplicates from Sorted array描述Given a sorted array, remove the duplicates in place such that each element appear only onceand return the new lengthDo not allocate extra space for another array, you must do this in place with constant memoryFor example, Given input array A =[1, 1, 2Your function should return length =2, and a is now [1, 2]分析无代码1/ LeetCode, Remove Duplicates from Sorted Array//时间复杂度0(n),空间复杂度0(1)class Solution tublicint removeDuplicates(vector& nums)tif (nums empty o) return 0;int index =ofor (int i =1: i nums size: 1++ iif (nums [index] ! nums [i])nums [++index]= nums [i]return index 12.1数组代码2//LeetCode, Remove Duplicates from Sorted Array/使用STL,时间复杂度0(n),空间复杂度0(1)class Solution ipublicint removeDuplicates(vector& nums)treturn distance(nums begin(), unique(nums begin(), nums end ())代码3/ LeetCode, Remove Duplicates from Sorted Array/使用STL,时间复杂度0(n),空间复杂度0(1)lass Solution fublicint removeDuplicates(vector& nums)treturn distance(nums begin(, removeDuplicates(nums begin(, nums end(), nums begintemplateOutIt removeDuplicates(InIt first, InIt last, OutIt output)thile (first last)i*output++ = *firstfirst upper_bound(first, last, *firstreturn output相关题目Remove Duplicates from Sorted Array I,见§2.1.22.1.2 Remove Duplicates from Sorted Array II描述Follow up for"Remove Duplicates " What if duplicates are allowed at most twice?For example, Given sorted array a =[1, 1, 1, 2, 2, 3]Your function should return length=5, and A is now [1, 1, 2, 2, 3分析加一个变量记录一下元素出现的次数即可。这题因为是已经排序的数组,所以一个变量即可解决。如果是没有排序的数组,则需要引入一个 hashmap来记录出现次数。4第2章线性表代码1// Leet Code, Remove Duplicates from Sorted Array II//时间复杂度0(n),空间复杂度0(1)//qauthorhex108(https://github.com/hex108)class Solution tublicint removeDuplicates(vector& nums)tif (nums size(
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