DIV1 250pt

题意：给定一个vector<int>A，若能从里面选出一些数，使得他们按位或的值为x，则称x为吉利数。给定k，问最少要从A里面去掉多少个数，才能使k变为不吉利数。

解法：按位考虑。如果A中某元素A[i]，将A[i]和k转化成二进制形式，如果某一位A[i]为1而k的为0，则一定不选选取掉A[i]，把所有这样的A[i]全部从A中删掉。然后，维护ans，枚举所有k二进制为1的位，计A中有t个元素该位为1，则ans = min(ans, t)。

A.size() <= 50，A[i] <= 10^9

tag：按位或

Ps：我的代码写的太复杂了。。。

 // BEGIN CUT HERE

 /*

  * Author:  plum rain

  * score :

  */

 /*

  */

 // END CUT HERE

 #line 11 "ORSolitaire.cpp"

 #include <sstream>

 #include <stdexcept>

 #include <functional>

 #include <iomanip>

 #include <numeric>

 #include <fstream>

 #include <cctype>

 #include <iostream>

 #include <cstdio>

 #include <vector>

 #include <cstring>

 #include <cmath>

 #include <algorithm>

 #include <cstdlib>

 #include <set>

 #include <queue>

 #include <bitset>

 #include <list>

 #include <string>

 #include <utility>

 #include <map>

 #include <ctime>

 #include <stack>

 using namespace std;

 #define CLR(x) memset(x, 0, sizeof(x))

 #define CLR1(x) memset(x, -1, sizeof(x))

 #define PB push_back

 #define SZ(v) ((int)(v).size())

 #define zero(x) (((x)>0?(x):-(x))<eps)

 #define out(x) cout<<#x<<":"<<(x)<<endl

 #define tst(a) cout<<#a<<endl

 #define CINBEQUICKER std::ios::sync_with_stdio(false)

 typedef vector<int> VI;

 typedef vector<string> VS;

 typedef vector<double> VD;

 typedef pair<int, int> pii;

 typedef long long int64;

 const double eps = 1e-;

 const double PI = atan(1.0)*;

 const int maxint = ;

 int num[];

 bool v[];

 bool ok(int x, int y)

 {

     while (y){

         int t1 = x & , t2 = y & ;

         if (!t2 && t1) return ;

         x >>= ; y >>= ;

     }

     return !x;

 }

 void gao(int x)

 {

     int p = ;

     while (x){

         num[p++] += (x & );

         x >>= ;

     }

 }

 class ORSolitaire

 {

     public:

         int getMinimum(vector <int> nb, int x){

             int sz = nb.size();

             CLR (num); CLR (v);

             for (int i = ; i < sz; ++ i){

                 v[i] = ok(nb[i], x);

                 if (v[i]) gao(nb[i]);

             }

             int p = ;

             int ans = ;

             while (x){

                 if (x & )

                     ans = min(ans, num[p]);

                 x >>= ;

                 ++ p;

             }

             return ans;

         }

 // BEGIN CUT HERE

     public:

     void run_test(int Case) { if ((Case == -) || (Case == )) test_case_0(); if ((Case == -) || (Case == )) test_case_1(); if ((Case == -) || (Case == )) test_case_2(); if ((Case == -) || (Case == )) test_case_3(); if ((Case == -) || (Case == )) test_case_4(); }

     private:

     template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }

     void verify_case(int Case, const int &Expected, const int &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }

     void test_case_0() { int Arr0[] = {, , }; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; int Arg2 = ; verify_case(, Arg2, getMinimum(Arg0, Arg1)); }

     void test_case_1() { int Arr0[] = {, , , , }; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; int Arg2 = ; verify_case(, Arg2, getMinimum(Arg0, Arg1)); }

     void test_case_2() { int Arr0[] = {, , }; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; int Arg2 = ; verify_case(, Arg2, getMinimum(Arg0, Arg1)); }

     void test_case_3() { int Arr0[] = {,,,,,,,,,,}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; int Arg2 = ; verify_case(, Arg2, getMinimum(Arg0, Arg1)); }

     void test_case_4() { int Arr0[] = {, , , , , , , , , , , , , , , , , , , }; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; int Arg2 = ; verify_case(, Arg2, getMinimum(Arg0, Arg1)); }

 // END CUT HERE

 };

 // BEGIN CUT HERE

 int main()

 {

 //    freopen( "a.out" , "w" , stdout );

     ORSolitaire ___test;

     ___test.run_test(-);

        return ;

 }

 // END CUT HERE

DIV1 600pt

题意：有一个矩阵每位只可能为1或0，每次操作可以将某一为的0变1或者1变0，求最少操作次数，使得操作以后，该矩阵有rnum行为回文，有cnum列为回文。

矩阵行数 <= 14，列数 <= 14，行数列数均为偶数。

解法：枚举行，dp列。用O(2^14)的复杂度枚举哪些行为对称行。处理列的时候，把关于中间对称的两列一起处理，每对列都可以看成4种物品，两列都不对称，左列对称右列不对称，右列对称左列不对称，都对称，然后用背包处理就行了。

tag:dp, 背包, good

 // BEGIN CUT HERE

 /*

 */

 // END CUT HERE

 #line 7 "PalindromeMatrix.cpp"

 #include <cstdlib>

 #include <cctype>

 #include <cstring>

 #include <cstdio>

 #include <cmath>

 #include <algorithm>

 #include <vector>

 #include <iostream>

 #include <sstream>

 #include <set>

 #include <queue>

 #include <fstream>

 #include <numeric>

 #include <iomanip>

 #include <bitset>

 #include <list>

 #include <stdexcept>

 #include <functional>

 #include <string>

 #include <utility>

 #include <map>

 #include <ctime>

 #include <stack>

 using namespace std;

 #define clr0(x) memset(x, 0, sizeof(x))

 #define clr1(x) memset(x, -1, sizeof(x))

 #define PB push_back

 #define sz(v) ((int)(v).size())

 #define out(x) cout<<#x<<":"<<(x)<<endl

 #define tst(a) cout<<#a<<endl

 #define CINBEQUICKER std::ios::sync_with_stdio(false)

 typedef vector<int> VI;

 typedef vector<string> VS;

 typedef vector<double> VD;

 typedef long long int64;

 const double eps = 1e-;

 const double PI = atan(1.0)*;

 const int maxint = ;

 inline int MyMod( int a , int b ) { return (a%b+b)%b;}

 VS a;

 bool v[];

 int d[][];

 int gao(int sta, int n, int m, int cnum)

 {

     int nmid = (n - ) >> , mmid = (m - ) >> ;

     clr0 (v);

     for (int i = ; i < n; ++ i)

         if (sta & ( << i)) v[i] = ;

     int c[][];

     for (int i = ; i <= mmid; ++ i){

         clr0 (c[i]);

         for (int j = ; j <= nmid; ++ j){

             int ii = m -  - i, jj = n -  - j;

             int t0 = ;

             if (a[jj][i] == '') ++ t0;

             if (a[jj][ii] == '') ++  t0;

             if (a[j][i] == '') ++ t0;

             if (a[j][ii] == '') ++ t0;

             if (v[j] && v[jj]){

                 c[i][] += (a[j][i] != a[j][ii]) + (a[jj][i] != a[jj][ii]);

                 int tmp = min(t0,  - t0);

                 for (int j = ; j < ; ++ j) c[i][j] += tmp;

             }

             if (v[j] && !v[jj]){

                 c[i][] += (a[j][i] != a[j][ii]);

                 if (a[j][i] == a[j][ii]){

                     c[i][] += (a[j][i] != a[jj][i]);

                     c[i][] += (a[j][i] != a[jj][ii]);

                 }

                 else{

                     ++ c[i][]; ++ c[i][];

                 }

                 c[i][] += min(t0, -t0);

             }

             if (!v[j] && v[jj]){

                 c[i][] += (a[jj][i] != a[jj][ii]);

                 if (a[jj][i] == a[jj][ii]){

                     c[i][] += (a[j][i] != a[jj][ii]);

                     c[i][] += (a[j][ii] != a[jj][ii]);

                 }

                 else{

                     ++ c[i][]; ++ c[i][];

                 }

                 c[i][] += min(t0, -t0);

             }

             if (!v[j] && !v[jj]){

                 int t1 = (a[j][i] != a[jj][i]), t2 = (a[j][ii] != a[jj][ii]);

                 c[i][] += t1; c[i][] += t2; c[i][] += t1 + t2;

             }

         }

     }

     clr1 (d);

     d[][] = c[][]; d[][] = min(c[][], c[][]); d[][] = c[][];

     for (int i = ; i <= mmid; ++ i)

         for (int j = ; j <= m; ++ j){

             if (d[i-][j] >= ) d[i][j] = d[i-][j] + c[i][];

             if (j && d[i-][j-] >= ) d[i][j] = min(d[i][j]>= ? d[i][j] : n*m+, d[i-][j-] + min(c[i][], c[i][]));

             if (j >=  && d[i-][j-] >= ) d[i][j] = min(d[i][j]>= ? d[i][j] : n*m+, d[i-][j-] + c[i][]);

         }

     int ret = n * m + ;

     for (int i = cnum; i <= m; ++ i){

         if (d[mmid][i] == -) continue;

         ret = min (ret, d[mmid][i]);

     }

     return ret;

 }

 class PalindromeMatrix

 {

     public:

         int minChange(vector <string> A, int rnum, int cnum){

             a = A;

             int n = sz(a), m = sz(a[]);

             int ans = n * m + , cnt =  << n;

             for (int i = ; i < cnt; ++ i)

                 if (__builtin_popcount(i) >= rnum) ans = min(ans, gao(i, n, m, cnum));

             return ans;

         }

 // BEGIN CUT HERE

     public:

     void run_test(int Case) { if ((Case == -) || (Case == )) test_case_0(); if ((Case == -) || (Case == )) test_case_1(); if ((Case == -) || (Case == )) test_case_2(); if ((Case == -) || (Case == )) test_case_3(); if ((Case == -) || (Case == )) test_case_4(); if ((Case == -) || (Case == )) test_case_5(); if ((Case == -) || (Case == )) test_case_6(); }

     //void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0();}

     private:

     template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }

     void verify_case(int Case, const int &Expected, const int &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }

     void test_case_0() { string Arr0[] = {""

 ,""

 ,""

 ,""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; int Arg2 = ; int Arg3 = ; verify_case(, Arg3, minChange(Arg0, Arg1, Arg2)); }

     void test_case_1() { string Arr0[] = {""

 ,""

 ,""

 ,""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; int Arg2 = ; int Arg3 = ; verify_case(, Arg3, minChange(Arg0, Arg1, Arg2)); }

     void test_case_2() { string Arr0[] = {""

 ,""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; int Arg2 = ; int Arg3 = ; verify_case(, Arg3, minChange(Arg0, Arg1, Arg2)); }

     void test_case_3() { string Arr0[] = {""

 ,""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; int Arg2 = ; int Arg3 = ; verify_case(, Arg3, minChange(Arg0, Arg1, Arg2)); }

     void test_case_4() { string Arr0[] = {""

 ,""

 ,""

 ,""

 ,""

 ,""

 ,""

 ,""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; int Arg2 = ; int Arg3 = ; verify_case(, Arg3, minChange(Arg0, Arg1, Arg2)); }

     void test_case_5() { string Arr0[] = {""

 ,""

 ,""

 ,""

 ,""

 ,""

 ,""

 ,""

 ,""

 ,""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; int Arg2 = ; int Arg3 = ; verify_case(, Arg3, minChange(Arg0, Arg1, Arg2)); }

     void test_case_6() { string Arr0[] = {""

 ,""

 ,""

 ,""

 ,""

 ,""

 ,""

 ,""

 ,""

 ,""

 ,""

 ,""

 ,""

 ,""}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; int Arg2 = ; int Arg3 = ; verify_case(, Arg3, minChange(Arg0, Arg1, Arg2)); }

 // END CUT HERE

 };

 //by plum rain

 // BEGIN CUT HERE

 int main()

 {

     //freopen( "a.out" , "w" , stdout );

     PalindromeMatrix ___test;

     ___test.run_test(-);

        return ;

 }

 // END CUT HERE

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