Instead of finding the longest common
subsequence, let us try to determine the
length of the LCS.
Then tracking back to find the LCS.
Consider a1a2…am and b1b2…bn.
Case 1: am=bn. The LCS must contain am,
we have to find the LCS of a1a2…am-1 and
b1b2…bn-1.
Case 2: am≠bn. Wehave to find the LCS of
a1a2…am-1 and b1b2…bn, and a1a2…am and
b b b
b1b2…bn-1
Let A = a1 a2 … am and B = b1 b2 … bn
Let Li j denote the length of the longest i,g g
common subsequence of a1 a2 … ai and b1 b2
… bj.
Li,j = Li-1,j-1 + 1 if ai=bj
max{ L L } a≠b i-1,j, i,j-1 if ai≠j
L0,0 = L0,j = Li,0 = 0 for 1≤i≤m, 1≤j≤n.
標簽:
the
subsequence
determine
Instead
上傳時間:
2013-12-17
上傳用戶:evil
MIPS 仿真器,能夠實現匯編,反匯編和模擬運行。自己開發的,java課的大程,可能問題比較多,僅做參考,而且程序架構也設計地不是很好。但作為新手大程學習和簡單的MIPS模擬還是沒有問題的。
本 MIPS 模擬器支持的指令如下:
add,addi,sub,subi,and,andi,or,ori,nor,beq,bne,j,jr,jal,lw,lh,lb,sw,sh,sb, 其中所有的跳
轉指令第三個操作數只能為一個行標簽[不支持相對地址以及絕對地址],標簽可
以寫在一行開頭,以冒號結尾。
標簽:
MIPS
仿真器
上傳時間:
2014-01-04
上傳用戶:ruixue198909