UVa Problem Solution: 861 - Little Bishops

 

Imagine that a chessboard with black and white squares. If a bishop is on a white square, can it attack bishops on black squares? No. So you can divide the chessboard into two independent ones. Then rotate each board for PI/4 and now the diagonal line becomes vertical. Now the problem of bishops becomes that of rooks.

 

We denote the number of cells of the i'th row contains with r[i]. In fact, the order of rows does not matter, so we can sort the rows with r[i]. We use c[i][j] to represent the number of cases to put j rooks in the first i rows. The we have the following recurrence,

      c[i][j] = c[i-1][j] + c[i-1][j-1] * (r[i] - (j-1))

with the boundary conditions,

      c[i][0] = 1, 0 <= i <= n, and

      c[0][j] = 0, 1 <= j <= k

 

Now, we have got the results of the two boards. A simple accumulation would give us the final results.

 

Code:

/*************************************************************************** * Copyright (C) 2008 by Liu Kaipeng * * LiuKaipeng at gmail dot com * ***************************************************************************/ /* @JUDGE_ID 00000 861 C++ "Little Bishops" */ #include <algorithm> #include <cstdio> #include <cstring> #include <deque> #include <fstream> #include <iostream> #include <list> #include <map> #include <queue> #include <set> #include <stack> #include <string> #include <vector> using namespace std; long long bishops(int n, int k) { long long r1[9]; long long r2[9]; for (int i = 1; i <= n; ++i) r1[i] = i % 2 != 0 ? i : r1[i-1]; for (int i = 1; i <= n-1; ++i) r2[i] = i % 2 != 0 ? i+1 : r2[i-1]; long long c1[9][65] = {​{0}}; long long c2[9][65] = {​{0}}; for (int i = 0; i <= n; ++i) c1[i][0] = 1; for (int j = 1; j <= k; ++j) c1[0][j] = 0; for (int i = 1; i <= n; ++i) for (int j = 1; j <= k && j <= i; ++j) c1[i][j] = c1[i-1][j] + c1[i-1][j-1]*(r1[i] - j + 1); for (int i = 0; i <= n-1; ++i) c2[i][0] = 1; for (int j = 1; j <= k; ++j) c2[0][j] = 0; for (int i = 1; i <= n-1; ++i) for (int j = 1; j <= k && j <= i; ++j) c2[i][j] = c2[i-1][j] + c2[i-1][j-1]*(r2[i] - j + 1); long long r = 0; for (int i = 0; i <= k; ++i) r += c1[n][i] * c2[n-1][k-i]; return r; } int main(int argc, char *argv[]) { #ifndef ONLINE_JUDGE freopen((string(argv[0]) + ".in").c_str(), "r", stdin); freopen((string(argv[0]) + ".out").c_str(), "w", stdout); #endif for (int n, k; cin >> n >> k && !(n == 0 && k == 0); ) cout << bishops(n, k) << endl; return 0; }