DLX的应用，基本题，注意maxnode开大点儿。
1 /* 3909 */
2 #include <iostream>
3 #include <string>
4 #include <map>
5 #include <queue>
6 #include <set>
7 #include <stack>
8 #include <vector>
9 #include <deque>
10 #include <algorithm>
11 #include <cstdio>
12 #include <cmath>
13 #include <ctime>
14 #include <cstring>
15 #include <climits>
16 #include <cctype>
17 #include <cassert>
18 #include <functional>
19 #include <iterator>
20 #include <iomanip>
21 using namespace std;
22 //#pragma comment(linker,"/STACK:102400000,1024000")
23
24 #define sti set<int>
25 #define stpii set<pair<int, int> >
26 #define mpii map<int,int>
27 #define vi vector<int>
28 #define pii pair<int,int>
29 #define vpii vector<pair<int,int> >
30 #define rep(i, a, n) for (int i=a;i<n;++i)
31 #define per(i, a, n) for (int i=n-1;i>=a;--i)
32 #define clr clear
33 #define pb push_back
34 #define mp make_pair
35 #define fir first
36 #define sec second
37 #define all(x) (x).begin(),(x).end()
38 #define SZ(x) ((int)(x).size())
39 #define lson l, mid, rt<<1
40 #define rson mid+1, r, rt<<1|1
41
42 typedef struct DLX {
43 static const int maxc = 4*16*16+5;
44 static const int maxr = 16*16*16+5;
45 static const int maxnode = 16*16*16*5+5;
46
47 int n, sz;
48 int S[maxc];
49
50 int row[maxnode], col[maxnode];
51 int L[maxnode], R[maxnode], U[maxnode], D[maxnode];
52
53 int ansd, cnt, ans[maxr], ans_[maxr];
54
55 void init(int n_) {
56 cnt = 0;
57 n = n_;
58
59 rep(i, 0, n+1) {
60 L[i] = i-1;
61 R[i] = i+1;
62 U[i] = i;
63 D[i] = i;
64 }
65
66 L[0] = n;
67 R[n] = 0;
68
69 sz = n+1;
70 memset(S, 0, sizeof(S));
71 }
72
73 void addRow(int r, vi columns) {
74 int first = sz;
75 int size = SZ(columns);
76
77 rep(i, 0, size) {
78 int c = columns[i];
79
80 L[sz] = sz-1;
81 R[sz] = sz+1;
82
83 D[sz] = c;
84 U[sz] = U[c];
85 D[U[c]] = sz;
86 U[c] = sz;
87
88 row[sz] = r;
89 col[sz] = c;
90
91 ++S[c];
92 ++sz;
93 }
94
95 R[sz - 1] = first;
96 L[first] = sz - 1;
97 }
98
99 void remove(int c) {
100 L[R[c]] = L[c];
101 R[L[c]] = R[c];
102 for (int i=D[c]; i!=c; i=D[i]) {
103 for (int j=R[i]; j!=i; j=R[j]) {
104 U[D[j]] = U[j];
105 D[U[j]] = D[j];
106 --S[col[j]];
107 }
108 }
109 }
110
111 void restore(int c) {
112 L[R[c]] = c;
113 R[L[c]] = c;
114 for (int i=D[c]; i!=c; i=D[i]) {
115 for (int j=R[i]; j!=i; j=R[j]) {
116 U[D[j]] = j;
117 D[U[j]] = j;
118 ++S[col[j]];
119 }
120 }
121 }
122
123 bool dfs(int d) {
124 if (R[0] == 0) {
125 ansd = d;
126 ++cnt;
127 rep(i, 0, ansd)
128 ans_[i] = ans[i];
129 return cnt>1;
130 }
131
132 int c = R[0];
133 for (int i=R[0]; i!=0; i=R[i]) {
134 if (S[i] < S[c])
135 c = i;
136 }
137
138 remove(c);
139 for (int i=D[c]; i!=c; i=D[i]) {
140 ans[d] = row[i];
141 for (int j=R[i]; j!=i; j=R[j]) {
142 remove(col[j]);
143 }
144 if (dfs(d + 1)) return true;
145 for (int j=L[i]; j!=i; j=L[j]) {
146 restore(col[j]);
147 }
148 }
149 restore(c);
150
151 return false;
152 }
153
154 void solve(vi& v) {
155
156 dfs(0);
157
158 if (cnt == 1) {
159 v.clr();
160 rep(i, 0, ansd)
161 v.pb(ans_[i]);
162 }
163 }
164
165 } DLX;
166
167 DLX solver;
168 int n, n2, n2n2;
169 const int maxl = 20;
170 char M[maxl][maxl], M_[maxl][maxl];
171 const int SLOT = 0;
172 const int ROW = 1;
173 const int COL = 2;
174 const int SUB = 3;
175
176 int encode(int a, int b, int c) {
177 return a*n2n2 + b*n2 + c + 1;
178 }
179
180 void decode(int code, int& a, int& b, int& c) {
181 --code;
182 c = code % n2;
183 code /= n2;
184 b = code % n2;
185 code /= n2;
186 a = code;
187 }
188
189 int getVal(char c) {
190 if (c>=‘0‘ && c<=‘9‘) return c-‘1‘;
191 return c-‘A‘+9;
192 }
193
194 int getChar(int val) {
195 if (val < 9) return val+‘1‘;
196 return val-9+‘A‘;
197 }
198
199 void init() {
200 n2 = n * n;
201 n2n2 = n2 * n2;
202 }
203
204 void solve(vi& ans) {
205 solver.init(4 * n2n2);
206
207 rep(r, 0, n2) {
208 rep(c, 0, n2) {
209 rep(v, 0, n2) {
210 if (M[r][c]==‘.‘ || v==getVal(M[r][c])) {
211 vi columns;
212 columns.pb(encode(SLOT, r, c));
213 columns.pb(encode(ROW, r, v));
214 columns.pb(encode(COL, c, v));
215 columns.pb(encode(SUB, r/n*n+c/n, v));
216 solver.addRow(encode(r, c, v), columns);
217 }
218 }
219 }
220 }
221
222 solver.solve(ans);
223 }
224
225 int main() {
226 ios::sync_with_stdio(false);
227 #ifndef ONLINE_JUDGE
228 freopen("data.in", "r", stdin);
229 freopen("data.out", "w", stdout);
230 #endif
231
232 vi ans, tmp;
233
234 while (scanf("%d", &n) != EOF) {
235 init();
236 rep(i, 0, n2)
237 scanf("%s", M[i]);
238
239 solve(ans);
240
241 if (solver.cnt == 0) {
242 puts("No Solution");
243 continue;
244 } else if (solver.cnt > 1) {
245 puts("Multiple Solutions");
246 continue;
247 }
248
249 // check if is minimal
250 bool flag = true;
251
252 rep(r, 0, n2) {
253 rep(c, 0, n2) {
254 if (M[r][c] != ‘.‘) {
255 char ch = M[r][c];
256 M[r][c] = ‘.‘;
257 solve(tmp);
258 M[r][c] = ch;
259 if (solver.cnt <= 1) {
260 flag = false;
261 goto _output;
262 }
263 }
264 }
265 }
266
267 _output:
268 if (flag) {
269 int sz = SZ(ans);
270 rep(i, 0, sz) {
271 int r, c, v;
272 decode(ans[i], r, c, v);
273 char ch = getChar(v);
274 M[r][c] = ch;
275 }
276
277 rep(i, 0, n2)
278 puts(M[i]);
279 } else {
280 puts("Not Minimal");
281 }
282 }
283
284 #ifndef ONLINE_JUDGE
285 printf("time = %d.\n", (int)clock());
286 #endif
287
288 return 0;
289 }
时间： 2024-12-28 00:45:08