Topic given
2 sequence, we want to seek LCS, but the length of the sequence is 250*250, the time complexity of the LCS is O (n*n), so we can not solve the number of the first sequence, by location, map to 1, 2, 3, 4, 5, 6, 7, 8, 9 then you will get a mapping function, the second sequence , mapping to a new sequence is equivalent to using a mapping function to map two strings to two new strings we will find that the length of the LCS will not change because the same character mappings in two sequences are still the same, so only two sequences of LCS are required and then we find that the first sequence is incremented, Then the LCS must also be incremented, so we're just asking for the longest increment subsequence of the second sequence. The answer is: Yes and Lis has O (NLOGN) algorithm so the complexity is reduced so that the problem can be solved
1#include <stdio.h>2#include <string.h>3#include <stdlib.h>4#include <algorithm>5#include <iostream>6#include <queue>7#include <stack>8#include <vector>9#include <map>Ten#include <Set> One#include <string> A using namespacestd; -typedefLong LongLL; - Const intINF =1<< -; the Const intN =90000; - intA[n],b[n]; - intReflect[n]; - intSt[n],top; + intMain () - { + intt,n,p,q,i,j,tcase; Ascanf"%d",&tCase); at for(t=1; t<=tcase; ++t) - { -top =0; -scanf"%d%d%d",&n,&p,&q); -p++; -q++; in for(i=1; i<=p; ++i) - { toscanf"%d",&a[i]); +Reflect[a[i]] =i; - } the for(i=1; i<=q; ++i) * { $scanf"%d",&b[i]);Panax NotoginsengB[i] =Reflect[b[i]]; - } theSt[top] = b[1]; + for(i=2; i<=q; ++i) A { the if(B[i] >St[top]) + { -St[++top] =B[i]; $ } $ Else - { - intLow =0, high =top; the while(Low <=High ) - {Wuyi intMid = (low + high) >>1; the if(B[i] >St[mid]) -Low = mid +1; Wu Else -High = mid-1; About } $St[low] =B[i]; - } - } -printf"Case %d:%d\n", t,top+1); A } + the return 0; -}
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HDU's Magic Board also uses a mapping, which becomes pre-processing and then reads the data to output the answer immediately.
uva10635 LCS mapping to Lis