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。。。
1 // BEGIN CUT HERE 2 /* 3 * Author: plum rain 4 * score : 5 */ 6 /* 7 8 */ 9 // END CUT HERE 10 #line 11 "ORSolitaire.cpp" 11 #include <sstream> 12 #include <stdexcept> 13 #include <functional> 14 #include <iomanip> 15 #include <numeric> 16 #include <fstream> 17 #include <cctype> 18 #include <iostream> 19 #include <cstdio> 20 #include <vector> 21 #include <cstring> 22 #include <cmath> 23 #include <algorithm> 24 #include <cstdlib> 25 #include <set> 26 #include <queue> 27 #include <bitset> 28 #include <list> 29 #include <string> 30 #include <utility> 31 #include <map> 32 #include <ctime> 33 #include <stack> 34 35 using namespace std; 36 37 #define CLR(x) memset(x, 0, sizeof(x)) 38 #define CLR1(x) memset(x, -1, sizeof(x)) 39 #define PB push_back 40 #define SZ(v) ((int)(v).size()) 41 #define zero(x) (((x)>0?(x):-(x))<eps) 42 #define out(x) cout<<#x<<":"<<(x)<<endl 43 #define tst(a) cout<<#a<<endl 44 #define CINBEQUICKER std::ios::sync_with_stdio(false) 45 46 typedef vector<int> VI; 47 typedef vector<string> VS; 48 typedef vector<double> VD; 49 typedef pair<int, int> pii; 50 typedef long long int64; 51 52 const double eps = 1e-8; 53 const double PI = atan(1.0)*4; 54 const int maxint = 2139062143; 55 56 int num[50]; 57 bool v[55]; 58 59 bool ok(int x, int y) 60 { 61 while (y){ 62 int t1 = x & 1, t2 = y & 1; 63 if (!t2 && t1) return 0; 64 x >>= 1; y >>= 1; 65 } 66 return !x; 67 } 68 69 void gao(int x) 70 { 71 int p = 0; 72 while (x){ 73 num[p++] += (x & 1); 74 x >>= 1; 75 } 76 } 77 78 class ORSolitaire 79 { 80 public: 81 int getMinimum(vector <int> nb, int x){ 82 int sz = nb.size(); 83 CLR (num); CLR (v); 84 for (int i = 0; i < sz; ++ i){ 85 v[i] = ok(nb[i], x); 86 if (v[i]) gao(nb[i]); 87 } 88 89 int p = 0; 90 int ans = 100; 91 while (x){ 92 if (x & 1) 93 ans = min(ans, num[p]); 94 x >>= 1; 95 ++ p; 96 } 97 return ans; 98 } 99 100 // BEGIN CUT HERE 101 public: 102 void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); if ((Case == -1) || (Case == 4)) test_case_4(); } 103 private: 104 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(); } 105 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 << " Expected: "" << Expected << '"' << endl; cerr << " Received: "" << Received << '"' << endl; } } 106 void test_case_0() { int Arr0[] = {1, 2, 4}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 7; int Arg2 = 1; verify_case(0, Arg2, getMinimum(Arg0, Arg1)); } 107 void test_case_1() { int Arr0[] = {1, 2, 4, 7, 8}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 7; int Arg2 = 2; verify_case(1, Arg2, getMinimum(Arg0, Arg1)); } 108 void test_case_2() { int Arr0[] = {12571295, 2174218, 2015120}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 1; int Arg2 = 0; verify_case(2, Arg2, getMinimum(Arg0, Arg1)); } 109 void test_case_3() { int Arr0[] = {5,2,4,52,62,9,8,3,1,11,6}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 11; int Arg2 = 3; verify_case(3, Arg2, getMinimum(Arg0, Arg1)); } 110 void test_case_4() { int Arr0[] = {503, 505, 152, 435, 491, 512, 1023, 355, 510, 500, 502, 255, 63, 508, 509, 511, 60, 250, 254, 346}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 510; int Arg2 = 5; verify_case(4, Arg2, getMinimum(Arg0, Arg1)); } 111 112 // END CUT HERE 113 114 }; 115 116 // BEGIN CUT HERE 117 int main() 118 { 119 // freopen( "a.out" , "w" , stdout ); 120 ORSolitaire ___test; 121 ___test.run_test(-1); 122 return 0; 123 } 124 // 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
1 // BEGIN CUT HERE 2 /* 3 4 */ 5 // END CUT HERE 6 #line 7 "PalindromeMatrix.cpp" 7 #include <cstdlib> 8 #include <cctype> 9 #include <cstring> 10 #include <cstdio> 11 #include <cmath> 12 #include <algorithm> 13 #include <vector> 14 #include <iostream> 15 #include <sstream> 16 #include <set> 17 #include <queue> 18 #include <fstream> 19 #include <numeric> 20 #include <iomanip> 21 #include <bitset> 22 #include <list> 23 #include <stdexcept> 24 #include <functional> 25 #include <string> 26 #include <utility> 27 #include <map> 28 #include <ctime> 29 #include <stack> 30 31 using namespace std; 32 33 #define clr0(x) memset(x, 0, sizeof(x)) 34 #define clr1(x) memset(x, -1, sizeof(x)) 35 #define PB push_back 36 #define sz(v) ((int)(v).size()) 37 #define out(x) cout<<#x<<":"<<(x)<<endl 38 #define tst(a) cout<<#a<<endl 39 #define CINBEQUICKER std::ios::sync_with_stdio(false) 40 41 typedef vector<int> VI; 42 typedef vector<string> VS; 43 typedef vector<double> VD; 44 typedef long long int64; 45 46 const double eps = 1e-8; 47 const double PI = atan(1.0)*4; 48 const int maxint = 2139062143; 49 50 inline int MyMod( int a , int b ) { return (a%b+b)%b;} 51 52 VS a; 53 bool v[20]; 54 int d[20][50]; 55 56 int gao(int sta, int n, int m, int cnum) 57 { 58 int nmid = (n - 1) >> 1, mmid = (m - 1) >> 1; 59 clr0 (v); 60 for (int i = 0; i < n; ++ i) 61 if (sta & (1 << i)) v[i] = 1; 62 63 int c[20][4]; 64 for (int i = 0; i <= mmid; ++ i){ 65 clr0 (c[i]); 66 for (int j = 0; j <= nmid; ++ j){ 67 int ii = m - 1 - i, jj = n - 1 - j; 68 int t0 = 0; 69 if (a[jj][i] == '0') ++ t0; 70 if (a[jj][ii] == '0') ++ t0; 71 if (a[j][i] == '0') ++ t0; 72 if (a[j][ii] == '0') ++ t0; 73 if (v[j] && v[jj]){ 74 c[i][0] += (a[j][i] != a[j][ii]) + (a[jj][i] != a[jj][ii]); 75 int tmp = min(t0, 4 - t0); 76 for (int j = 1; j < 4; ++ j) c[i][j] += tmp; 77 } 78 if (v[j] && !v[jj]){ 79 c[i][0] += (a[j][i] != a[j][ii]); 80 if (a[j][i] == a[j][ii]){ 81 c[i][1] += (a[j][i] != a[jj][i]); 82 c[i][2] += (a[j][i] != a[jj][ii]); 83 } 84 else{ 85 ++ c[i][1]; ++ c[i][2]; 86 } 87 c[i][3] += min(t0, 4-t0); 88 } 89 if (!v[j] && v[jj]){ 90 c[i][0] += (a[jj][i] != a[jj][ii]); 91 if (a[jj][i] == a[jj][ii]){ 92 c[i][1] += (a[j][i] != a[jj][ii]); 93 c[i][2] += (a[j][ii] != a[jj][ii]); 94 } 95 else{ 96 ++ c[i][1]; ++ c[i][2]; 97 } 98 c[i][3] += min(t0, 4-t0); 99 } 100 if (!v[j] && !v[jj]){ 101 int t1 = (a[j][i] != a[jj][i]), t2 = (a[j][ii] != a[jj][ii]); 102 c[i][1] += t1; c[i][2] += t2; c[i][3] += t1 + t2; 103 } 104 } 105 } 106 107 clr1 (d); 108 d[0][0] = c[0][0]; d[0][1] = min(c[0][1], c[0][2]); d[0][2] = c[0][3]; 109 for (int i = 1; i <= mmid; ++ i) 110 for (int j = 0; j <= m; ++ j){ 111 if (d[i-1][j] >= 0) d[i][j] = d[i-1][j] + c[i][0]; 112 if (j && d[i-1][j-1] >= 0) d[i][j] = min(d[i][j]>=0 ? d[i][j] : n*m+10, d[i-1][j-1] + min(c[i][1], c[i][2])); 113 if (j >= 2 && d[i-1][j-2] >= 0) d[i][j] = min(d[i][j]>=0 ? d[i][j] : n*m+10, d[i-1][j-2] + c[i][3]); 114 } 115 116 int ret = n * m + 10; 117 for (int i = cnum; i <= m; ++ i){ 118 if (d[mmid][i] == -1) continue; 119 ret = min (ret, d[mmid][i]); 120 } 121 return ret; 122 } 123 124 class PalindromeMatrix 125 { 126 public: 127 int minChange(vector <string> A, int rnum, int cnum){ 128 a = A; 129 int n = sz(a), m = sz(a[0]); 130 int ans = n * m + 10, cnt = 1 << n; 131 for (int i = 0; i < cnt; ++ i) 132 if (__builtin_popcount(i) >= rnum) ans = min(ans, gao(i, n, m, cnum)); 133 return ans; 134 } 135 136 // BEGIN CUT HERE 137 public: 138 void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); if ((Case == -1) || (Case == 4)) test_case_4(); if ((Case == -1) || (Case == 5)) test_case_5(); if ((Case == -1) || (Case == 6)) test_case_6(); } 139 //void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0();} 140 private: 141 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(); } 142 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 << " Expected: "" << Expected << '"' << endl; cerr << " Received: "" << Received << '"' << endl; } } 143 void test_case_0() { string Arr0[] = {"0000" 144 ,"1000" 145 ,"1100" 146 ,"1110"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 2; int Arg2 = 2; int Arg3 = 1; verify_case(0, Arg3, minChange(Arg0, Arg1, Arg2)); } 147 void test_case_1() { string Arr0[] = {"0000" 148 ,"1000" 149 ,"1100" 150 ,"1110"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 3; int Arg2 = 2; int Arg3 = 3; verify_case(1, Arg3, minChange(Arg0, Arg1, Arg2)); } 151 void test_case_2() { string Arr0[] = {"01" 152 ,"10"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 1; int Arg2 = 1; int Arg3 = 1; verify_case(2, Arg3, minChange(Arg0, Arg1, Arg2)); } 153 void test_case_3() { string Arr0[] = {"1110" 154 ,"0001"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 0; int Arg2 = 0; int Arg3 = 0; verify_case(3, Arg3, minChange(Arg0, Arg1, Arg2)); } 155 void test_case_4() { string Arr0[] = {"01010101" 156 ,"01010101" 157 ,"01010101" 158 ,"01010101" 159 ,"01010101" 160 ,"01010101" 161 ,"01010101" 162 ,"01010101"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 2; int Arg2 = 3; int Arg3 = 8; verify_case(4, Arg3, minChange(Arg0, Arg1, Arg2)); } 163 void test_case_5() { string Arr0[] = {"000000000000" 164 ,"011101110111" 165 ,"010001010101" 166 ,"010001010101" 167 ,"011101010101" 168 ,"010101010101" 169 ,"010101010101" 170 ,"011101110111" 171 ,"000000000000" 172 ,"000000000000"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 5; int Arg2 = 9; int Arg3 = 14; verify_case(5, Arg3, minChange(Arg0, Arg1, Arg2)); } 173 void test_case_6() { string Arr0[] = {"11111101001110" 174 ,"11000111111111" 175 ,"00010101111001" 176 ,"10110000111111" 177 ,"10000011010010" 178 ,"10001101101101" 179 ,"00101010000001" 180 ,"10111010100100" 181 ,"11010011110111" 182 ,"11100010110110" 183 ,"00100101010100" 184 ,"01001011001000" 185 ,"01010001111010" 186 ,"10100000010011"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 6; int Arg2 = 8; int Arg3 = 31; verify_case(6, Arg3, minChange(Arg0, Arg1, Arg2)); } 187 188 // END CUT HERE 189 190 }; 191 //by plum rain 192 // BEGIN CUT HERE 193 int main() 194 { 195 //freopen( "a.out" , "w" , stdout ); 196 PalindromeMatrix ___test; 197 ___test.run_test(-1); 198 return 0; 199 } 200 // END CUT HERE