*[topcoder]AstronomicalRecords

http://community.topcoder.com/stat?c=problem_statement&pm=12804&rd=15706

首先A和B的长度都不一定一样，里面的元素也不一定有序。比如，A={1,4,6,7,2,8} and B={1,3,5,6,8,2,9,10}。二来，因为A和B都是代表着ratio，那么选定A[i]和B[j]，假设他们是同一个，就可以把A和B数组normalize。方法是A*B[j]，B*A[i]。此时问题就转化成最长公共子序列了。另外要注意，因为用了乘法，所以元素可能超过int，要用long表示。

#include <set>

#include <vector>

using namespace std;



class AstronomicalRecords {

public:

    int minimalPlanets(vector <int> A, vector <int> B);

private:

    int sub(vector<long> A, vector<long> B, int i, int j);

};



int AstronomicalRecords::sub(vector<long> A, vector<long> B, int x, int y) {

    int p = A[x];

    int q = B[y];

    for (int i = 0; i < A.size(); i++) {

        A[i] *= q;

    }

    for (int i = 0; i < B.size(); i++) {

        B[i] *= p;

    }

    vector<vector<int> > dp(A.size()+1);

	for (int i = 0; i < dp.size(); i++)

		dp[i].resize(B.size()+1, 0); // i, j for the length

	int m = 0;

    for (int i = 0; i <= A.size(); i++) {

        for (int j = 0; j <= B.size(); j++) {

            if (i == 0 || j == 0)

                dp[i][j] = 0;

            else if (A[i-1] == B[j-1])

                dp[i][j] = dp[i-1][j-1] + 1;

            else

                dp[i][j] = dp[i-1][j-1];

			m = max(dp[i][j], m);

        }

    }

    return m;

}



int AstronomicalRecords::minimalPlanets(vector <int> _A, vector <int> _B) {

    vector<long> A;

    vector<long> B;

    for (int i = 0; i < _A.size(); i++) {

        A.push_back(_A[i]);

    }

    for (int i = 0; i < _B.size(); i++) {

        B.push_back(_B[i]);

    }

    int max_seq = 0;

    for (int i = 0; i < A.size(); i++) {

        for (int j = 0; j < B.size(); j++) {

            max_seq = max(sub(A, B, i, j), max_seq);

        }

    }

    return A.size() + B.size() - max_seq;

}