topcoder SRM 594 DIV2 AstronomicalRecordsEasy

此题主要考查的是求最长公共子序列

设A[i]:A[j] = a:b = ac:bc       B[ii]:B[jj] = c:d = ac:ad ,

如果A[i]:A[j] = B[ii]:B[jj]则bc= ad，所以A乘以B的倍数，B乘以A的倍数，则A与B所求的序列必然是一样的，即求A与B的最长公共子序列

 #include <iostream>
 #include <vector>
 #include <algorithm>

 using namespace std;

 class AstronomicalRecordsEasy{
 public:
     int minimalPlanets(vector<int> A, vector<int> B){
         int lenA = A.size(),lenB = B.size(),res = lenA + lenB;
         for(int i = ; i < lenA; ++ i ){
             for(int  j =  ; j < lenB; ++ j){
                 int a = A[i],b = B[j];
                 vector<vector<int> > dp(lenA+,vector<int> (lenB+,));
                 for(int ki = ; ki <= lenA; ++ ki ){
                     for(int kj = ; kj <= lenB; ++ kj){
                         dp[ki][kj] = max(max(dp[ki-][kj],dp[ki][kj-]),dp[ki-][kj-] + (b*A[ki-] == a*B[kj-]) );
                     }
                 }
                 res = min(res, lenA+lenB - dp[lenA][lenB]);
             }
         }
         return res;
     }
 };