LCS(最長公共子序列)問題可以簡單地描述如下: 一個給定序列的子序列是在該序列中刪去若干元素后得到的序列。給定兩個序列X和Y,當另一序列Z既是X的子序列又是Y的子序列時,稱Z是序列X和Y的公共子序列。例如,若X={A,B,C,B,D,B,A},Y={B,D,C,A,B,A},則序列{B,C,A}是X和Y的一個公共子序列,但它不是X和Y的一個最長公共子序列。序列{B,C,B,A}也是X和Y的一個公共子序列,它的長度為4,而且它是X和Y的一個最長公共子序列,因為X和Y沒有長度大于4的公共子序列。 最長公共子序列問題就是給定兩個序列X={x1,x2,...xm}和Y={y1,y2,...yn},找出X和Y的一個最長公共子序列。對于這個問題比較容易想到的算法是窮舉,對X的所有子序列,檢查它是否也是Y的子序列,從而確定它是否為X和Y的公共子序列,并且在檢查過程中記錄最長的公共子序列。X的所有子序列都檢查過后即可求出X和Y的最長公共子序列。X的每個子序列相應于下標集{1,2,...,m}的一個子集。因此,共有2^m個不同子序列,從而窮舉搜索法需要指數時間。
上傳時間: 2015-06-09
上傳用戶:氣溫達上千萬的
c語言版的多項式曲線擬合。 用最小二乘法進行曲線擬合. 用p-1 次多項式進行擬合,p<= 10 x,y 的第0個域x[0],y[0],沒有用,有效數據從x[1],y[1] 開始 nNodeNum,有效數據節點的個數。 b,為輸出的多項式系數,b[i] 為b[i-1]次項。b[0],沒有用。 b,有10個元素ok。
上傳時間: 2014-01-12
上傳用戶:變形金剛
高精度乘法基本思想和加法一樣。其基本流程如下: ①讀入被乘數s1,乘數s2 ②把s1、s2分成4位一段,轉成數值存在數組a,b中;記下a,b的長度k1,k2; ③i賦為b中的最低位; ④從b中取出第i位與a相乘,累加到另一數組c中;(注意:累加時錯開的位數應是多少位 ?) ⑤i:=i-1;檢測i值:小于k2則轉⑥,否則轉④ ⑥打印結果
上傳時間: 2015-08-16
上傳用戶:源弋弋
This ActiveX control and Demo will allow you to dock your toolbars/forms to a mdiform much in the way in which Visual C++ allows you to. 象Vc++一樣使用多文本窗體
標簽: toolbars ActiveX control mdiform
上傳時間: 2015-09-14
上傳用戶:asdfasdfd
ejb3.0 in action的全部配套源代碼,相信讀過spring in action,ajax in action的讀者應該已經領閱到action從書的魅力,這本書是最新的action從書,對于正在學習ejb3.0的同志無疑是最好的幫助,源代碼與書配套,相得益彰。
上傳時間: 2015-10-03
上傳用戶:woshiayin
In this first-ever paperback edition of his long-time best-seller, motivational speaker Steve Chandler helps you create an action plan for living your vision in business and in life. It features 100 proven methods to positively change the way you think and act-methods based on feedback from the hundreds of thousands of corporate and public seminar attendees Chandler speaks to each year. 100 Ways to Motivate Yourself will help you break through the negative barriers and banish the pessimistic thoughts that are preventing you from fulfilling your lifelong goals and dreams. Whether you re self-employed, a manager, or a high-level executive, it s still easy to get stuck in the daily routines of life, fantasizing about what could have been. Steve Chandler helps you turn that way of thinking around and make what could have been into what can and will be.
標簽: motivational best-seller first-ever paperback
上傳時間: 2015-10-26
上傳用戶:牛津鞋
[輸入] 圖的頂點個數N,圖中頂點之間的關系及起點A和終點B [輸出] 若A到B無路徑,則輸出“There is no path” 否則輸出A到B路徑上個頂點 [存儲結構] 圖采用鄰接矩陣的方式存儲。 [算法的基本思想] 采用廣度優先搜索的方法,從頂點A開始,依次訪問與A鄰接的頂點VA1,VA2,...,VAK, 訪問遍之后,若沒有訪問B,則繼續訪問與VA1鄰接的頂點VA11,VA12,...,VA1M,再訪問與VA2鄰接頂點...,如此下去,直至找到B,最先到達B點的路徑,一定是邊數最少的路徑。實現時采用隊列記錄被訪問過的頂點。每次訪問與隊頭頂點相鄰接的頂點,然后將隊頭頂點從隊列中刪去。若隊空,則說明到不存在通路。在訪問頂點過程中,每次把當前頂點的序號作為與其鄰接的未訪問的頂點的前驅頂點記錄下來,以便輸出時回溯。 #include<stdio.h> int number //隊列類型 typedef struct{ int q[20]
標簽: 輸入
上傳時間: 2015-11-16
上傳用戶:ma1301115706
The initial planning and thinking about this book began during a discussion of SQL Server futures in July 2001. The discussion was with Rob Howard during a trip to Microsoft to discuss the first book I was working on at that time. After that, I stayed involved in what was happening in ADO.NET by going to the SQL Server Yukon Technical Preview in Bellevue, Washington, in February 2002 and by working with the ASP.NET and SQL Server teams at Microsoft since July 2003.
標簽: discussion planning thinking initial
上傳時間: 2014-01-08
上傳用戶:cjf0304
the calculator s usage! after you have inputed 2 operators,choose + - * / function! But the only situation I did t deal with is that when you choos + fuction ,and the operaters signs is like this -A+B,just turn it to B-A!
標簽: calculator the operators function
上傳時間: 2016-02-12
上傳用戶:lili123
CRC16算法的Java實現,使用方法如下: CRC16 crc16 = new CRC16() byte[] b = new byte[] { // (byte) 0xF0,(byte)0xF0,(byte)0xF0,(byte)0x72 (byte) 0x2C, (byte) 0x00, (byte) 0xFF, (byte) 0xFE, (byte) 0xFE, (byte) 0x04, (byte) 0x00, (byte) 0x00, (byte) 0x00, (byte) 0x00 } for (int k = 0 k < b.length k++) { crc16.update(b[k]) } System.out.println(Integer.toHexString(crc16.getValue())) System.out.println(Integer.toHexString(b.length))
上傳時間: 2014-12-20
上傳用戶:ve3344