1. 题目描述
由$n \times n, n \in [1, 5]$的正方形由$2 \times n \times (n+1)$根木棍组成，可能已经有些木棍被破坏，求至少还需破坏多少木根，可以使得不存在任何正方形？
2. 基本思路
这是一道非常有趣的题目，可以使用IDA*解也可以用DLX解。可以试试5 0这组数据比较两者的性能差异。
(1) IDA*
使用IDA*处理，是因为最后的解的范围一定不是很大，因为数据很小。至多也就60根木棍。
首先可以预处理分别对正方形和木棍进行编号，进而预处理破坏每根木棍可以影响的正方形。
由于木棍和正方形都不超过60个，因此可以采用longlong进行状态压缩，一定要装压，否则就哭吧。
每次选择当前仍然存在的最小正方形，然后枚举它的每条边进行深搜。
(2) DLX
使用DLX解这道题很容易理解， 而且解5 0这组数据也很快。
首先仍然可以使用上述的预处理方法。从而可以得到当前木棍中仍然存在的正方形。
如果，当前不存在正方形则直接输出0。否则以正方形的个数初始化DLX。
接下来，枚举每条边，搜索它可以影响到的正方形，建立一个可行方案。
直接套用DLX可解，注意DLX加入H函数可以加速，而且效果还是挺明显的。
3. 代码
(1)IDA*
1 /* 1084 */
2 #include <iostream>
3 #include <sstream>
4 #include <string>
5 #include <map>
6 #include <queue>
7 #include <set>
8 #include <stack>
9 #include <vector>
10 #include <deque>
11 #include <bitset>
12 #include <algorithm>
13 #include <cstdio>
14 #include <cmath>
15 #include <ctime>
16 #include <cstring>
17 #include <climits>
18 #include <cctype>
19 #include <cassert>
20 #include <functional>
21 #include <iterator>
22 #include <iomanip>
23 using namespace std;
24 //#pragma comment(linker,"/STACK:102400000,1024000")
25
26 #define sti set<int>
27 #define stpii set<pair<int, int> >
28 #define mpii map<int,int>
29 #define vi vector<int>
30 #define pii pair<int,int>
31 #define vpii vector<pair<int,int> >
32 #define rep(i, a, n) for (int i=a;i<n;++i)
33 #define per(i, a, n) for (int i=n-1;i>=a;--i)
34 #define clr clear
35 #define pb push_back
36 #define mp make_pair
37 #define fir first
38 #define sec second
39 #define all(x) (x).begin(),(x).end()
40 #define SZ(x) ((int)(x).size())
41 #define lson l, mid, rt<<1
42 #define rson mid+1, r, rt<<1|1
43 #define INF 0x3f3f3f3f
44 #define mset(a, val) memset(a, (val), sizeof(a))
45
46 #define LL long long
47
48 typedef struct {
49 vector<LL> sq;
50 int sz;
51 } sqInfo_t;
52
53
54 const int maxm = 65;
55 int rowId[6][6];
56 int colId[6][6];
57 int n, tot;
58 LL st;
59 sqInfo_t sqInfo[6];
60
61 void init_sq(int n) {
62 int cnt = 0;
63 sqInfo_t& info = sqInfo[n];
64 vector<LL>& sq = info.sq;
65 int& sz = info.sz;
66
67 rep(i, 0, n) {
68 rep(j, 0, n)
69 rowId[i][j] = cnt++;
70 rep(j, 0, n+1)
71 colId[i][j] = cnt++;
72 }
73 rep(j, 0, n)
74 rowId[n][j] = cnt++;
75
76 cnt = 0;
77 rep(i, 0, n) {
78 rep(j, 0, n) {
79 rep(k, 1, n+1) {
80 if (i+k>n || j+k>n)
81 break;
82
83 LL val = 0;
84
85 // Top
86 rep(l, 0, k) {
87 val |= (1LL << rowId[i][j+l]);
88 }
89
90 // Left
91 rep(l, 0, k) {
92 val |= (1LL << colId[i+l][j]);
93 }
94
95 // Down
96 rep(l, 0, k) {
97 val |= (1LL << rowId[i+k][j+l]);
98 }
99
100 // Right
101 rep(l, 0, k) {
102 val |= (1LL << colId[i+l][j+k]);
103 }
104
105 sq.pb(val);
106 ++sz;
107 }
108 }
109 }
110 }
111
112 void init() {
113 rep(i, 1, 6)
114 init_sq(i);
115 }
116
117
118 bool dfs(int m, LL cst) {
119 const int& sz = sqInfo[n].sz;
120 const vector<LL>& sq = sqInfo[n].sq;
121 int h = 0;
122 LL st = -1, tmp = cst;
123
124 rep(i, 0, sz) {
125 if ((tmp & sq[i]) == sq[i]) {
126 ++h;
127 tmp ^= sq[i];
128 if (st < 0)
129 st = sq[i];
130 }
131 }
132
133 if(h == 0)
134 return true;
135
136 if (m < h)
137 return false;
138
139 rep(i, 0, tot) {
140 if (st & (1LL << i)) {
141 if ( dfs(m-1, cst^(1LL<<i)) ) return true;
142 }
143 }
144 return false;
145 }
146
147 void solve() {
148 int ans = -1;
149
150 for (int i=0; i<=tot; ++i) {
151 if (dfs(i, st)) {
152 ans = i;
153 break;
154 }
155 }
156
157 printf("%d\n", ans);
158 }
159
160 int main() {
161 ios::sync_with_stdio(false);
162 #ifndef ONLINE_JUDGE
163 freopen("data.in", "r", stdin);
164 freopen("data.out", "w", stdout);
165 #endif
166
167 int t;
168
169 init();
170 scanf("%d", &t);
171 while (t--) {
172 scanf("%d", &n);
173 tot = n*(n+1)*2;
174 int k, x;
175 st = (1LL << tot) - 1;
176
177 scanf("%d", &k);
178 while (k--) {
179 scanf("%d", &x);
180 --x;
181 st ^= (1LL << x);
182 }
183 solve();
184 }
185
186 #ifndef ONLINE_JUDGE
187 printf("time = %d.\n", (int)clock());
188 #endif
189
190 return 0;
191 }
(2)Dancing Links
1 /* 1084 */
2 #include <iostream>
3 #include <sstream>
4 #include <string>
5 #include <map>
6 #include <queue>
7 #include <set>
8 #include <stack>
9 #include <vector>
10 #include <deque>
11 #include <bitset>
12 #include <algorithm>
13 #include <cstdio>
14 #include <cmath>
15 #include <ctime>
16 #include <cstring>
17 #include <climits>
18 #include <cctype>
19 #include <cassert>
20 #include <functional>
21 #include <iterator>
22 #include <iomanip>
23 using namespace std;
24 //#pragma comment(linker,"/STACK:102400000,1024000")
25
26 #define sti set<int>
27 #define stpii set<pair<int, int> >
28 #define mpii map<int,int>
29 #define vi vector<int>
30 #define pii pair<int,int>
31 #define vpii vector<pair<int,int> >
32 #define rep(i, a, n) for (int i=a;i<n;++i)
33 #define per(i, a, n) for (int i=n-1;i>=a;--i)
34 #define clr clear
35 #define pb push_back
36 #define mp make_pair
37 #define fir first
38 #define sec second
39 #define all(x) (x).begin(),(x).end()
40 #define SZ(x) ((int)(x).size())
41 #define lson l, mid, rt<<1
42 #define rson mid+1, r, rt<<1|1
43 #define INF 0x3f3f3f3f
44 #define mset(a, val) memset(a, (val), sizeof(a))
45
46 #define LL __int64
47
48 typedef struct {
49 static const int maxr = 65;
50 static const int maxc = 65;
51 static const int maxn = maxr * maxc;
52
53 int n, sz;
54 int S[maxc];
55 bool visit[maxc];
56
57 int col[maxn];
58 int L[maxn], R[maxn], U[maxn], D[maxn];
59
60 int ansd;
61
62 void init(int _n) {
63 n = _n;
64
65 rep(i, 0, n+1) {
66 L[i] = i - 1;
67 R[i] = i + 1;
68 U[i] = i;
69 D[i] = i;
70 col[i] = i;
71 }
72
73 L[0] = n;
74 R[n] = 0;
75
76 sz = n + 1;
77 memset(S, 0, sizeof(S));
78 ansd = INF;
79 }
80
81 void addRow(const vi& columns) {
82 int first = sz;
83 int size = SZ(columns);
84
85 rep(i, 0, size) {
86 const int& c = columns[i];
87
88 L[sz] = sz - 1;
89 R[sz] = sz + 1;
90
91 D[sz] = c;
92 U[sz] = U[c];
93 D[U[c]] = sz;
94 U[c] = sz;
95
96 col[sz] = c;
97
98 ++S[c];
99 ++sz;
100 }
101
102 L[first] = sz - 1;
103 R[sz-1] = first;
104 }
105
106 void remove_col(int c) {
107 for (int i=D[c]; i!=c; i=D[i]) {
108 L[R[i]] = L[i];
109 R[L[i]] = R[i];
110 --S[col[i]];
111 }
112 }
113
114 void restore_col(int c) {
115 for (int i=D[c]; i!=c; i=D[i]) {
116 L[R[i]] = i;
117 R[L[i]] = i;
118 ++S[col[i]];
119 }
120 }
121
122 int H() {
123 int ret = 0;
124
125 memset(visit, false, sizeof(visit));
126 for (int i=R[0]; i; i=R[i]) {
127 if (visit[col[i]])
128 continue;
129
130 ++ret;
131 visit[col[i]] = true;
132 for (int j=D[i]; j!=i; j=D[j]) {
133 for (int k=R[j]; k!=j; k=R[k]) {
134 visit[col[k]] = true;
135 }
136 }
137 }
138
139 return ret;
140 }
141
142 void dfs(int d) {
143 int delta = H();
144
145 if (d+delta >= ansd)
146 return ;
147
148 if (R[0] == 0) {
149 ansd = d;
150 return ;
151 }
152
153 int c = R[0];
154 for (int i=R[0]; i; i=R[i]) {
155 if (S[i] < S[c])
156 c = i;
157 }
158
159 for (int i=D[c]; i!=c; i=D[i]) {
160 remove_col(i);
161 for (int j=R[i]; j!=i; j=R[j]) {
162 remove_col(j);
163 }
164 dfs(d + 1);
165 for (int j=L[i]; j!=i; j=L[j]) {
166 restore_col(j);
167 }
168 restore_col(i);
169 }
170 }
171
172 } DLX;
173
174 typedef struct {
175 vector<LL> sq;
176 int sz;
177 } sqInfo_t;
178
179 DLX solver;
180 const int maxm = 65;
181 bool mark[maxm];
182 int valid[maxm];
183 int rowId[6][6];
184 int colId[6][6];
185 int n, tot;
186 LL st;
187 sqInfo_t sqInfo[6];
188
189 void init_sq(int n) {
190 int cnt = 0;
191 sqInfo_t& info = sqInfo[n];
192 vector<LL>& sq = info.sq;
193 int& sz = info.sz;
194
195 rep(i, 0, n) {
196 rep(j, 0, n)
197 rowId[i][j] = cnt++;
198 rep(j, 0, n+1)
199 colId[i][j] = cnt++;
200 }
201 rep(j, 0, n)
202 rowId[n][j] = cnt++;
203
204 cnt = 0;
205 rep(i, 0, n) {
206 rep(j, 0, n) {
207 rep(k, 1, n+1) {
208 if (i+k>n || j+k>n)
209 break;
210
211 LL val = 0;
212
213 // Top
214 rep(l, 0, k) {
215 val |= (1LL << rowId[i][j+l]);
216 }
217
218 // Left
219 rep(l, 0, k) {
220 val |= (1LL << colId[i+l][j]);
221 }
222
223 // Down
224 rep(l, 0, k) {
225 val |= (1LL << rowId[i+k][j+l]);
226 }
227
228 // Right
229 rep(l, 0, k) {
230 val |= (1LL << colId[i+l][j+k]);
231 }
232
233 sq.pb(val);
234 ++sz;
235 }
236 }
237 }
238 }
239
240 void init() {
241 rep(i, 1, 6)
242 init_sq(i);
243 }
244
245 void solve() {
246 const vector<LL>& sq = sqInfo[n].sq;
247 const int& sz = sqInfo[n].sz;
248 int m = 0;
249
250 rep(i, 0, sz) {
251 if ((st & sq[i]) == sq[i]) {
252 valid[m++] = i;
253 }
254 }
255
256 if (m == 0) {
257 puts("0");
258 return ;
259 }
260
261 solver.init(m);
262
263 rep(i, 0, tot) {
264 if (mark[i]) {
265 vi vc;
266
267 rep(j, 0, m) {
268 if (sq[valid[j]] & (1LL << i))
269 vc.pb(j+1);
270 }
271
272 if (SZ(vc) > 0) {
273 solver.addRow(vc);
274 }
275 }
276 }
277
278 solver.dfs(0);
279 int ans = solver.ansd;
280
281 printf("%d\n", ans);
282 }
283
284 int main() {
285 ios::sync_with_stdio(false);
286 #ifndef ONLINE_JUDGE
287 freopen("data.in", "r", stdin);
288 freopen("data.out", "w", stdout);
289 #endif
290
291 int t;
292
293 init();
294 scanf("%d", &t);
295 while (t--) {
296 scanf("%d", &n);
297 memset(mark, true, sizeof(mark));
298 tot = n*(n+1)*2;
299 int k, x;
300 st = (1LL << tot) - 1;
301
302 scanf("%d", &k);
303 while (k--) {
304 scanf("%d", &x);
305 mark[--x] = false;
306 st ^= (1LL << x);
307 }
308 solve();
309 }
310
311 #ifndef ONLINE_JUDGE
312 printf("time = %d.\n", (int)clock());
313 #endif
314
315 return 0;
316 }
4. 数据生成器
1 import sys
2 import string
3 from random import randint, shuffle
4
5
6 def GenData(fileName):
7 with open(fileName, "w") as fout:
8 t = 20
9 fout.write("%d\n" % (t))
10 for tt in xrange(t):
11 n = randint(1, 5)
12 tot = n * (n + 1)
13 taken = randint(1, tot)
14 L = range(1, tot+1)
15 shuffle(L)
16 fout.write("%d %d\n" % (n, taken))
17 fout.write(" ".join(map(str, L[:taken])) + "\n")
18
19
20 def MovData(srcFileName, desFileName):
21 with open(srcFileName, "r") as fin:
22 lines = fin.readlines()
23 with open(desFileName, "w") as fout:
24 fout.write("".join(lines))
25
26
27 def CompData():
28 print "comp"
29 srcFileName = "F:\Qt_prj\hdoj\data.out"
30 desFileName = "F:\workspace\cpp_hdoj\data.out"
31 srcLines = []
32 desLines = []
33 with open(srcFileName, "r") as fin:
34 srcLines = fin.readlines()
35 with open(desFileName, "r") as fin:
36 desLines = fin.readlines()
37 n = min(len(srcLines), len(desLines))-1
38 for i in xrange(n):
39 ans2 = int(desLines[i])
40 ans1 = int(srcLines[i])
41 if ans1 > ans2:
42 print "%d: wrong" % i
43
44
45 if __name__ == "__main__":
46 srcFileName = "F:\Qt_prj\hdoj\data.in"
47 desFileName = "F:\workspace\cpp_hdoj\data.in"
48 GenData(srcFileName)
49 MovData(srcFileName, desFileName)
时间： 2024-10-03 20:42:40