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1 | 1 | ---
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2 |
| -title: Codeforces Round \#554 (Div. 2) |
| 2 | +title: Codeforces Round #554 (Div. 2) |
3 | 3 | categories:
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4 | 4 | - ACM
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5 | 5 | - 题解
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@@ -234,3 +234,89 @@ int main()
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234 | 234 | printf("%d ", rk[g.e[g.p[n - 2]].second]);
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235 | 235 | }
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236 | 236 | ```
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| 237 | +# [Neko Rules the Catniverse (Small Version)](https://vjudge.net/problem/CodeForces-1152F1) |
| 238 | +状压DP,用$f[i][j][mask]$来表示正在考虑第i个,已经考虑了第$j$个,后$m$个的选择状态是二进制位集$mask$。 |
| 239 | + |
| 240 | +这里可以滚动数组。 |
| 241 | +```cpp |
| 242 | +#include <bits/stdc++.h> |
| 243 | +using namespace std; |
| 244 | +void add(int &a, long long b, int M = 1e9 + 7) { a = (a + b) % M; } |
| 245 | +int n, k, m; |
| 246 | +int main() |
| 247 | +{ |
| 248 | + scanf("%d%d%d", &n, &k, &m); |
| 249 | + vector<vector<int>> f(k + 1, vector<int>(1 << m, 0)); |
| 250 | + f[0][0] = 1; |
| 251 | + for (int i = 0; i < n; ++i) |
| 252 | + { |
| 253 | + vector<vector<int>> g(k + 1, vector<int>(1 << m, 0)); |
| 254 | + for (int j = 0; j < f.size(); ++j) |
| 255 | + for (int mask = 0; mask < f[j].size(); ++mask) |
| 256 | + { |
| 257 | + int newMask = (mask << 1) % f[j].size(); |
| 258 | + add(g[j][newMask], f[j][mask]); |
| 259 | + if (j < k) |
| 260 | + add(g[j + 1][newMask | 1], (1LL + __builtin_popcount(mask)) * f[j][mask]); |
| 261 | + } |
| 262 | + f.swap(g); |
| 263 | + } |
| 264 | + for (int mask = m = 0; mask < f[k].size(); ++mask) |
| 265 | + add(m, f[k][mask]); |
| 266 | + printf("%d", m); |
| 267 | +} |
| 268 | +``` |
| 269 | +# [Neko Rules the Catniverse (Large Version)](https://vjudge.net/problem/CodeForces-1152F2) |
| 270 | +注意到状态只在相邻的两位转移,这里可以用矩阵加速DP。 |
| 271 | +
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| 272 | +下面这段代码跑了5070ms,官方还有一个[跑了156ms的解法](https://codeforces.com/contest/1152/submission/53260183),暂时没看懂…也太强了吧。 |
| 273 | +```cpp |
| 274 | +#include <bits/stdc++.h> |
| 275 | +using namespace std; |
| 276 | +int n, k, m; |
| 277 | +void add(int &a, long long b, int M = 1e9 + 7) { a = (a + b) % M; } |
| 278 | +int toId(int j, int mask) { return j << m | mask; } |
| 279 | +struct Matrix |
| 280 | +{ |
| 281 | + vector<vector<int>> a; |
| 282 | + int n; |
| 283 | + Matrix(int n) : n(n), a(n, vector<int>(n)) {} |
| 284 | + friend Matrix operator*(const Matrix &a, const Matrix &b) |
| 285 | + { |
| 286 | + Matrix c(a.n); |
| 287 | + for (int i = 0; i < a.n; ++i) |
| 288 | + for (int j = 0; j < a.n; ++j) |
| 289 | + for (int k = 0; k < a.n; ++k) |
| 290 | + add(c.a[i][j], 1LL * a.a[i][k] * b.a[k][j]); |
| 291 | + return c; |
| 292 | + } |
| 293 | + friend Matrix pow(Matrix a, int b) |
| 294 | + { |
| 295 | + Matrix r(a.n); |
| 296 | + for (int i = 0; i < a.n; ++i) |
| 297 | + r.a[i][i] = 1; |
| 298 | + for (; b; b >>= 1, a = a * a) |
| 299 | + if (b & 1) |
| 300 | + r = a * r; |
| 301 | + return r; |
| 302 | + } |
| 303 | +}; |
| 304 | +int main() |
| 305 | +{ |
| 306 | + scanf("%d%d%d", &n, &k, &m); |
| 307 | + Matrix dp(toId(k + 1, 0)), f(dp); |
| 308 | + for (int j = 0; j <= k; ++j) |
| 309 | + for (int mask = 0; mask < (1 << m); ++mask) |
| 310 | + { |
| 311 | + int newMask = (mask << 1) % (1 << m); |
| 312 | + f.a[toId(j, newMask)][toId(j, mask)] = 1; |
| 313 | + if (j < k) |
| 314 | + f.a[toId(j + 1, newMask | 1)][toId(j, mask)] = 1 + __builtin_popcount(mask); |
| 315 | + } |
| 316 | + dp.a[0][0] = 1, dp = pow(f, n) * dp; |
| 317 | + int ans = 0; |
| 318 | + for (int mask = 0; mask < (1 << m); ++mask) |
| 319 | + add(ans, dp.a[toId(k, mask)][0]); |
| 320 | + printf("%d", ans); |
| 321 | +} |
| 322 | +``` |
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