Sgu 223. Little Kings

Time Limit: 0.25 s

Space limit: 4 m

Question:

Place K (k <= N * n) kings (eight adjacent grids can be attacked) on the N * n (n ≤ 10) chessboard ), number of solutions that make them unable to attack each other.

 

 

 

 

 

 

Solution:

State compression is used, and K-bit (k <= N) binary number 1 is used to represent the position of the king placed on the board.

First, use preprocessing to find out the solution that meets the requirements when statu [I] only considers one row. C [I] is the number of pieces placed in the statu [I] State.

 

F [I] [J] [p] indicates the number of solutions for placing J pawns when row I is put in the P state.

 

State transition equation: F [I] [J] [p] = Σ f [I-1] [J-C [J] [PP], (the PP and P meeting the I-1 line do not conflict)

 

Determine if the conflict between P and PP is met

(P & PP) = 0 & (P <1 & PP) = 0 & (PP <1 & P) = 0

 

Finally, the output ans = Σ f [N] [k] [Pi]

 

Reference code:

 

 

#include <iostream>#include <cstdio>#define LL long longusing namespace std;int powT[11];int statu[1 << 10], c[1 << 10];LL f[11][111][1 << 10];int main() {int n, k, tol = 0, t;scanf ("%d %d", &n, &k);for (int i = 0; i <= ( (1 << n) - 1); i++) {if ( ( (i & (i >> 1) ) == 0) && ( (i & (i << 1) ) == 0) ) {statu[++tol] = i;for (t = i; t > 0; t >>= 1)if (t & 1) c[i]++;}}for (int i = 1; i <= tol; i++)f[1][c[statu[i]]][statu[i]] = 1;for (int i = 2; i <= n; i++) {for (int j = 0; j <= k; j++)for (int pu = 1; pu <= tol; pu++)for (int pv = 1; pv <= tol; pv++) {int p1 = statu[pu], p2 = statu[pv];if (j >= c[p1] && (p1 & p2) == 0 && ( p1 << 1 & p2) == 0 && ( p2 << 1 & p1) == 0)f[i][j][p1] += f[i - 1][j - c[p1]][p2];}}LL ans = 0;for (int i = 1; i <= tol; i++)ans += f[n][k][statu[i]];printf ("%lld", ans);}