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 }
时间： 2024-11-05 06:42:42