[csu1605]数独（精确覆盖问题）
题意 ：给定数独的某些初始值，规定每个格子的得分，求得分最大的数独的解。
思路：这是某年的noip的原题，高中时就写过，位运算也就是那个时候学会的--。这题明显是暴搜，但是需要注意两点，一是需要加一些常数优化，也就是位运算，一个是剪枝，填完某个数后发现某个格子无解了则换个数填，并且那些可填的数的种数少的格子尽量先填，因为这样尽可能让矛盾在靠近根的地方出现。今天粗略学了一下舞蹈链--DLX，这个算法（准确来说是一个结构）可以比较高效的解决一些精确覆盖问题，对于重复覆盖问题稍作修改也适用。用DLX写了一遍数独，发现效率比位运算略高一点，但不明显。
位运算：
1 #pragma comment(linker, "/STACK:10240000,10240000") 2 3 #include <iostream> 4 #include <cstdio> 5 #include <algorithm> 6 #include <cstdlib> 7 #include <cstring> 8 #include <map> 9 #include <queue> 10 #include <deque> 11 #include <cmath> 12 #include <vector> 13 #include <ctime> 14 #include <cctype> 15 #include <set> 16 #include <bitset> 17 #include <functional> 18 #include <numeric> 19 #include <stdexcept> 20 #include <utility> 21 22 using namespace std; 23 24 #define mem0(a) memset(a, 0, sizeof(a)) 25 #define mem_1(a) memset(a, -1, sizeof(a)) 26 #define lson l, m, rt << 1 27 #define rson m + 1, r, rt << 1 | 1 28 #define define_m int m = (l + r) >> 1 29 #define rep_up0(a, b) for (int a = 0; a < (b); a++) 30 #define rep_up1(a, b) for (int a = 1; a <= (b); a++) 31 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--) 32 #define rep_down1(a, b) for (int a = b; a > 0; a--) 33 #define all(a) (a).begin(), (a).end() 34 #define lowbit(x) ((x) & (-(x))) 35 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {} 36 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {} 37 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {} 38 #define pchr(a) putchar(a) 39 #define pstr(a) printf("%s", a) 40 #define sstr(a) scanf("%s", a) 41 #define sint(a) scanf("%d", &a) 42 #define sint2(a, b) scanf("%d%d", &a, &b) 43 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c) 44 #define pint(a) printf("%d\n", a) 45 #define test_print1(a) cout << "var1 = " << a << endl 46 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl 47 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl 48 49 typedef long long LL; 50 typedef pair<int, int> pii; 51 typedef vector<int> vi; 52 53 const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1}; 54 const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 }; 55 const int maxn = 3e4 + 7; 56 const int md = 10007; 57 const int inf = 1e9 + 7; 58 const LL inf_L = 1e18 + 7; 59 const double pi = acos(-1.0); 60 const double eps = 1e-6; 61 62 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);} 63 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;} 64 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;} 65 template<class T>T condition(bool f, T a, T b){return f?a:b;} 66 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];} 67 int make_id(int x, int y, int n) { return x * n + y; } 68 69 int ans, a[10][10], f[1 << 13], row[10], col[10], block[10], sp[1 << 13]; 70 71 int getScore(int i, int j) { 72 return min(min(i, 8 - i), min(j, 8 - j)) + 6; 73 } 74 75 void init() { 76 rep_up0(i, 12) { 77 f[1 << i] = i; 78 } 79 } 80 81 void dfs(int k, int score) { 82 if (k >= 81) { 83 max_update(ans, score); 84 return ; 85 } 86 int x, y, c = 10; 87 rep_up0(i, 9) { 88 bool ok = false; 89 rep_up0(j, 9) { 90 if (a[i][j]) continue; 91 int tmp = row[i] | col[j] | block[make_id(i / 3, j / 3, 3)]; 92 int tot = 0x3fe ^ tmp; 93 int cnt = 0; 94 if (tot == 0) { 95 ok = true; 96 c = 0; 97 break; 98 } 99 cnt = sp[tot]; 100 if (cnt < c) { 101 x = i; 102 y = j; 103 c = cnt; 104 } 105 } 106 if (ok) break; 107 } 108 if (c == 0 || c == 10) return ; 109 int i = x, j = y; 110 int tmp = row[i] | col[j] | block[make_id(i / 3, j / 3, 3)]; 111 int tot = 0x3fe ^ tmp; 112 while (tot) { 113 tmp = lowbit(tot); 114 row[i] ^= 1 << f[tmp]; 115 col[j] ^= 1 << f[tmp]; 116 block[make_id(i / 3, j / 3, 3)] ^= 1 << f[tmp]; 117 a[i][j] = f[tmp]; 118 dfs(k + 1, score + f[tmp] * getScore(i, j)); 119 row[i] ^= 1 << f[tmp]; 120 col[j] ^= 1 << f[tmp]; 121 block[make_id(i / 3, j / 3, 3)] ^= 1 << f[tmp]; 122 a[i][j] = 0; 123 tot -= tmp; 124 } 125 } 126 127 int main() { 128 //freopen("in.txt", "r", stdin); 129 sp[0] = 0; 130 rep_up1(i, 1 << 10) { 131 sp[i] = sp[i - lowbit(i)] + 1; 132 } 133 int T; 134 init(); 135 cin >> T; 136 while (T --) { 137 int sum = 0, cnt = 0, ok = true; 138 mem0(col); 139 mem0(row); 140 mem0(block); 141 rep_up0(i, 9) { 142 rep_up0(j, 9) { 143 sint(a[i][j]); 144 sum += a[i][j] * getScore(i, j); 145 if (a[i][j]) { 146 cnt ++; 147 if (col[j] & (1 << a[i][j])) ok = false; 148 if (row[i] & (1 << a[i][j])) ok = false; 149 if (block[make_id(i / 3, j / 3, 3)] & (1 << a[i][j])) ok = false; 150 col[j] |= 1 << a[i][j]; 151 row[i] |= 1 << a[i][j]; 152 block[make_id(i / 3, j / 3, 3)] |= 1 << a[i][j]; 153 } 154 } 155 } 156 ans = -1; 157 if (ok) dfs(cnt, sum); 158 cout << ans << endl; 159 } 160 }
DLX（模板）：
1 #pragma comment(linker, "/STACK:102400000,102400000") 2 3 #include <iostream> 4 #include <cstdio> 5 #include <algorithm> 6 #include <cstdlib> 7 #include <cstring> 8 #include <map> 9 #include <queue> 10 #include <deque> 11 #include <cmath> 12 #include <vector> 13 #include <ctime> 14 #include <cctype> 15 #include <set> 16 #include <bitset> 17 #include <functional> 18 #include <numeric> 19 #include <stdexcept> 20 #include <utility> 21 22 using namespace std; 23 24 #define mem0(a) memset(a, 0, sizeof(a)) 25 #define mem_1(a) memset(a, -1, sizeof(a)) 26 #define lson l, m, rt << 1 27 #define rson m + 1, r, rt << 1 | 1 28 #define define_m int m = (l + r) >> 1 29 #define rep_up0(a, b) for (int a = 0; a < (b); a++) 30 #define rep_up1(a, b) for (int a = 1; a <= (b); a++) 31 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--) 32 #define rep_down1(a, b) for (int a = b; a > 0; a--) 33 #define all(a) (a).begin(), (a).end() 34 #define lowbit(x) ((x) & (-(x))) 35 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {} 36 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {} 37 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {} 38 #define pchr(a) putchar(a) 39 #define pstr(a) printf("%s", a) 40 #define sstr(a) scanf("%s", a) 41 #define sint(a) scanf("%d", &a) 42 #define sint2(a, b) scanf("%d%d", &a, &b) 43 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c) 44 #define pint(a) printf("%d\n", a) 45 #define test_print1(a) cout << "var1 = " << a << endl 46 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl 47 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl 48 49 typedef long long LL; 50 typedef pair<int, int> pii; 51 typedef vector<int> vi; 52 53 const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1}; 54 const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 }; 55 const int maxn = 1e5 + 7; 56 const int md = 10007; 57 const int inf = 1e9 + 7; 58 const LL inf_L = 1e18 + 7; 59 const double pi = acos(-1.0); 60 const double eps = 1e-6; 61 62 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);} 63 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;} 64 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;} 65 template<class T>T condition(bool f, T a, T b){return f?a:b;} 66 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];} 67 int make_id(int x, int y, int n) { return x * n + y; } 68 69 ///行编号从1开始，列编号1~n，结点0是表头结点，结点1~n是各列顶部的虚拟结点 70 int result; 71 int b[10][10]; 72 73 int encode(int a, int b, int c) { 74 return a * 81 + b * 9 + c + 1; 75 } 76 void decode(int code, int &a, int &b, int &c) { 77 code --; 78 c = code % 9; code /= 9; 79 b = code % 9; code /= 9; 80 a = code; 81 } 82 83 struct DLX 84 { 85 const static int maxn = 1050; 86 const static int maxnode = 100007; 87 int n , sz; // 行数，节点总数 88 int S[maxn]; // 各列节点总数 89 int row[maxnode],col[maxnode]; // 各节点行列编号 90 int L[maxnode],R[maxnode],U[maxnode],D[maxnode]; // 十字链表 91 92 int ansd,ans[maxn]; // 解 93 94 void init(int n ) 95 { 96 this->n = n ; 97 for(int i = 0 ; i <= n; i++ ) 98 { 99 U[i] = i ; 100 D[i] = i ; 101 L[i] = i - 1; 102 R[i] = i + 1; 103 } 104 R[n] = 0 ; 105 L[0] = n; 106 sz = n + 1 ; 107 memset(S,0,sizeof(S)); 108 } 109 void addRow(int r,vector<int> c1) 110 { 111 int first = sz; 112 for(int i = 0 ; i < c1.size(); i++ ){ 113 int c = c1[i]; 114 L[sz] = sz - 1 ; R[sz] = sz + 1 ; D[sz] = c ; U[sz] = U[c]; 115 D[U[c]] = sz; U[c] = sz; 116 row[sz] = r; col[sz] = c; 117 S[c] ++ ; sz ++ ; 118 } 119 R[sz - 1] = first ; L[first] = sz - 1; 120 } 121 // 顺着链表A，遍历除s外的其他元素 122 #define FOR(i,A,s) for(int i = A[s]; i != s ; i = A[i]) 123 124 void remove(int c) { 125 L[R[c]] = L[c]; 126 R[L[c]] = R[c]; 127 FOR(i,D,c) 128 FOR(j,R,i) {U[D[j]] = U[j];D[U[j]] = D[j];--S[col[j]];} 129 } 130 void restore(int c) { 131 FOR(i,U,c) 132 FOR(j,L,i) {++S[col[j]];U[D[j]] = j;D[U[j]] = j; } 133 L[R[c]] = c; 134 R[L[c]] = c; 135 } 136 void update() { 137 int score = 0; 138 rep_up0(i, ansd) { 139 int r, c, v; 140 decode(ans[i], r, c, v); 141 score += (v + 1) * b[r][c]; 142 } 143 max_update(result, score); 144 } 145 bool dfs(int d) { 146 if(R[0] == 0) { 147 ansd = d; 148 update(); 149 return true; 150 } 151 // 找S最小的列c 152 int c = R[0]; 153 FOR(i,R,0) if(S[i] < S[c]) c = i; 154 155 remove(c); 156 FOR(i,D,c) { 157 ans[d] = row[i]; 158 FOR(j,R,i) remove(col[j]); 159 //if(dfs(d + 1)) return true; 160 dfs(d + 1); 161 FOR(j,L,i) restore(col[j]); 162 } 163 restore(c); 164 165 //return false; 166 } 167 bool solve(vector<int> & v) { 168 v.clear(); 169 if(!dfs(0)) return false; 170 for(int i = 0 ; i < ansd ;i ++) v.push_back(ans[i]); 171 return true; 172 } 173 }; 174 175 DLX solver; 176 int a[12][12]; 177 178 179 int main() { 180 //freopen("in.txt", "r", stdin); 181 rep_up0(i, 9) { 182 rep_up0(j, 9) { 183 b[i][j] = 6 + min(min(i, 8 - i), min(j, 8 - j)); 184 } 185 } 186 int T, x; 187 cin >> T; 188 while (T --) { 189 solver.init(324); 190 rep_up0(i, 9) { 191 rep_up0(j, 9) { 192 int x; 193 sint(x); 194 rep_up0(k, 9) { 195 if (x == 0 || x == k + 1) { 196 vector<int> col; 197 col.push_back(encode(0, i, j)); 198 col.push_back(encode(1, i, k)); 199 col.push_back(encode(2, j, k)); 200 col.push_back(encode(3, make_id(i / 3, j / 3, 3), k)); 201 solver.addRow(encode(i, j, k), col); 202 } 203 } 204 } 205 } 206 result = -1; 207 solver.dfs(0); 208 cout << result << endl; 209 } 210 return 0; 211 }
