topcoder SRM 594 DIV2 AstronomicalRecordsEasy

此題主要考查的是求最長公共子序列

設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]則bc= ad，所以A乘以B的倍數，B乘以A的倍數，則A與B所求的序列必然是一樣的，即求A與B的最長公共子序列

#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 = 0; i < lenA; ++ i ){
            for(int  j = 0 ; j < lenB; ++ j){
                int a = A[i],b = B[j];
                vector<vector<int> > dp(lenA+1,vector<int> (lenB+1,0));
                for(int ki = 1; ki <= lenA; ++ ki ){
                    for(int kj = 1; kj <= lenB; ++ kj){
                        dp[ki][kj] = max(max(dp[ki-1][kj],dp[ki][kj-1]),dp[ki-1][kj-1] + (b*A[ki-1] == a*B[kj-1]) );
                    }
                }
                res = min(res, lenA+lenB - dp[lenA][lenB]);
            }
        }
        return res;
    }
};