[Bzoj 2004] [Hnoi2010] Bus bus line "pressure dp + matrix multiplication"

Title Link: BZOJ-2004

Problem analysis

See the topic is not at all. He immediately looked at the Ben of God's men.

Because of the p<=10, I think it is using the pressure. Each successive P-position is compressed into a P-bit 2 binary number, in which a total of k bits is 1, indicating that the K-position is the current stop position of a bus. It is important to note that the first bit of each State must be 1, so that there is no duplicate status. Each state can be transferred to some state on the right (moved by the first 1 of the current state). Both the initial and terminating states are preceded by a K-bit of 1. Transfer n-k times with matrices.

Code

#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath > #include <algorithm>using namespace std;const int maxmap = 5, Mod = 30031;int N, K, P, Top, Rec;int L[MAXM Ap];int Calc (int Num) {int Cnt = 0;while (num) {++cnt; num-= num &-num;} return Cnt;} BOOL Check (int x, int y) {x = (x-(1 << (P-1))) << 1;int t = x ^ y;if (t-(t &-t) = = 0) return True;r Eturn false;} struct Matrix {int x, y, num[maxmap][maxmap];void setxy (int a, int b) {x = A; y = b;} void Clear (int xx) {for (int i = 1, i <= x; ++i) {for (int j = 1; J <= y; ++j) {num[i][j] = xx;}}} M0, MZ; Matrix Mul (Matrix A, Matrix B) {matrix Ret;ret. SetXY (a.x, B.Y); ret. Clear (0), for (int i = 1, i <= ret.x; ++i) {for (int j = 1, J <= Ret.y; ++j) {for (int k = 1; k <= a.y; ++k) {ret. NUM[I][J] + = a.num[i][k] * B.num[k][j];ret. NUM[I][J]%= Mod;}}} return ret;} Matrix Pow (Matrix A, int b) {matrix ret, f;f = A;ret. SetXY (f.x, F.Y); ret. Clear(0); for (int i = 1; I <= ret.x; ++i) ret. Num[i][i] = 1;while (b) {if (b & 1) ret = Mul (ret, f); b >>= 1;f = Mul (f, f);} return ret;} int main () {scanf ("%d%d%d", &n, &k, &p); Top = 0;for (int i = (1 << (P-1)), I <= (1 << P)-1; ++i) {if (Calc (i) = = K) {L[++top] = i; if (i = = (1 << P)-1-((1 << (P-K))-1) Rec = Top; }}mz. SetXY (top, top); Mz. Clear (0); M0. SetXY (1, TOP); M0. Clear (0); M0. Num[1][rec] = 1;for (int i = 1; I <= top; ++i) {for (int j = 1; j <= Top; ++j) {if (Check (L[i], l[j]) MZ. NUM[I][J] = 1;}} MZ = Pow (mz, n-k); M0 = Mul (M0, MZ);p rintf ("%d\n", M0. NUM[1][REC]); return 0;}

　　

[Bzoj 2004] [Hnoi2010] Bus bus line "pressure dp + matrix multiplication"