[状压 + 矩阵乘法] HDU 4332 Constructing Chimney

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题目大意：我们计划建一个大烟囱。烟囱的每一节是一个 3*3 的正方形，中间是空的，如图所示。现在我们只有许多 1*1*2 的块。我们建造的烟囱的高度为 N。现在请计算有多少种不同的建造烟囱的方法。答案模 10^9 +7。

据说是很经典的一道状压矩乘 dp 。

首先可以对前三层裸状压 dp 求出状态到状态的转移。可以发现，有用的状态只有 70 个。然后写到矩乘里面即可。

记得开 long long，还有 Case 的首字母要大写！！！

#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#define mo 1000000007LLint t, n, m;int num[71];int a[256][256], b[256][256], c[256][256], (* f)[256];bool v1[256], v2[256], v3[256], * vst;struct ma{long long a[71][71];long long * operator [](const int & b) {return a[b];}}p, q, x, y;ma operator *(ma & a, ma & b){ma c = ma();for (int i = 1; i <= 70; ++ i)for (int j = 1; j <= 70; ++ j)for (int k = 1; k <= 70; ++ k)c[i][j] = (c[i][j] + a[i][k] * b[k][j]) % mo;return c;}ma operator /(ma & a, ma & b){ma c = ma();for (int j = 1; j <= 70; ++ j)for (int k = 1; k <= 70; ++ k)c[1][j] = (c[1][j] + a[1][k] * b[k][j]) % mo;return c;}void show(int t, int k){if (t) show(t >> 1, k);fprintf(stderr, "%d", t & 1);if (t == k) fprintf(stderr, "\n");}void dfs(int r, int s, int t, int l){if (r == (1 << 8) - 1) return (void)(++ f[s][t], vst[t] = 1);for (int i = l; i < 7; ++ i)if (! (r >> i & 1)){dfs(r | 1 << i, s, t | 1 << i, i + 1);if (! (r >> i + 1 & 1)) dfs(r | 1 << i | 1 << i + 1, s, t, i + 2);}if (! (r >> 7 & 1)){dfs(r | 1 << 7, s, t | 1 << 7, 8);if (! (r & 1)) dfs(r | 1 << 7 | 1, s, t, 9);}}void prepare(){f = a, vst = v1, dfs(0, 0, 0, 0);f = b, vst = v2;for (int i = 0; i < 1 << 8; ++ i)if (v1[i]) dfs(i, i, 0, 0);f = c, vst = v3;for (int i = 0; i < 1 << 8; ++ i)if (v2[i]) dfs(i, i, 0, 0), num[++ m] = i;for (int i = 1; i <= 70; ++ i){p[1][i] = a[0][num[i]];for (int j = 1; j <= 70; ++ j) q[j][i] = c[num[i]][num[j]];}}int main(){freopen("a.in", "r", stdin);freopen("a.out", "w", stdout);prepare(), scanf("%d", & t);for (int i = 1; i <= t; ++ i){scanf("%d", & n), x = p, y = q;for (-- n; n; n >>= 1, y = y * y)if (n & 1) x = x / y;printf("Case %d: %I64d\n", i, x[1][1]);}return 0;}