Fairy Chess
Problem Statement :
Let's play Fairy Chess! You have an n*n chessboard. An s-leaper is a chess piece which can move from some square (x0,y0) to some square (x1,y1) if abs(x0-x1) + abs(y0-y1) <= s; however, its movements are restricted to up, down, left, and right within the confines of the chessboard, meaning that diagonal moves are not allowed. In addition, the leaper cannot leap to any square that is occupied by a pawn. Given the layout of the chessboard, can you determine the number of ways a leaper can move m times within the chessboard? Note: abs(x) refers to the absolute value of some integer, x. Input Format The first line contains an integer, q, denoting the number of queries. Each query is described as follows: 1.The first line contains three space-separated integers denoting n, m, and s, respectively. 2.Each line i of the n subsequent lines contains n characters. The jth character in the ith line describes the contents of square (i,j) according to the following key: . indicates the location is empty. P indicates the location is occupied by a pawn. L indicates the location of the leaper. Constraints 1 <= q <= 10 1 <= m <= 200 There will be exactly one L character on the chessboard. The s-leaper can move up (), down (), left (), and right () within the confines of the chessboard. It cannot move diagonally.
Solution :
Solution in C :
In C++ :
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cassert>
using namespace std;
const int P = 1000000007;
const int maxN = 200;
const int maxS = 200;
int TN, TC;
int N, M, S, K;
char tmp[maxN + 1];
char board[maxN][maxN];
int sx, sy;
int num[maxN][maxN];
int sum[2][maxN][maxN];
#define PLUS(x, v) \
{ \
x += v; \
if (x >= P) \
x -= P; \
}
#define MINUS(x, v) \
{ \
x -= v; \
if (x < 0) \
x += P; \
}
void find_leaper (int &px, int &py)
{
for (int x = 0; x < N; ++x)
for (int y = 0; y < N; ++y)
if (board[x][y] == 'L')
{
px = x;
py = y;
return;
}
}
void coor_trans (int x, int y, int &tx, int &ty, int &s)
{
if (!((x + y) & 1))
{
s = 0;
tx = (x + y) >> 1;
ty = ((y - x) >> 1) + ((N - 1) >> 1);
}
else
{
s = 1;
tx = (x + y) >> 1;
ty = ((y - x + 1) >> 1) + ((N - 2) >> 1);
}
}
int square_sum (int x, int y, int h)
{
int tx, ty, s;
coor_trans(x, y, tx, ty, s);
int tx2 = tx - h, ty2 = ty - h;
if (tx >= N)
tx = N - 1;
if (ty >= N)
ty = N - 1;
if (tx2 >= N)
tx2 = N - 1;
if (ty2 >= N)
ty2 = N - 1;
int r = sum[s][tx][ty];
if (tx2 >= 0)
MINUS(r, sum[s][tx2][ty]);
if (ty2 >= 0)
MINUS(r, sum[s][tx][ty2]);
if (tx2 >= 0 && ty2 >= 0)
PLUS(r, sum[s][tx2][ty2]);
return r;
}
int solve ()
{
for (int x = 0; x < N; ++x)
for (int y = 0; y < N; ++y)
if (board[x][y] != 'P')
num[x][y] = 1;
else
num[x][y] = 0;
for (int u = 0; u < M; ++u)
{
memset(sum, 0, sizeof(sum));
for (int x = 0; x < N; ++x)
for (int y = 0; y < N; ++y)
if (board[x][y] != 'P')
{
int tx, ty, s;
coor_trans(x, y, tx, ty, s);
sum[s][tx][ty] = num[x][y];
}
for (int s = 0; s < 2; ++s)
{
for (int y = 1; y < N; ++y)
PLUS(sum[s][0][y], sum[s][0][y - 1]);
for (int x = 1; x < N; ++x)
{
PLUS(sum[s][x][0], sum[s][x - 1][0]);
for (int y = 1; y < N; ++y)
{
int &r = sum[s][x][y];
PLUS(r, sum[s][x - 1][y]);
PLUS(r, sum[s][x][y - 1]);
MINUS(r, sum[s][x - 1][y - 1]);
}
}
}
for (int x = 0; x < N; ++x)
for (int y = 0; y < N; ++y)
if (board[x][y] != 'P')
{
num[x][y] = square_sum(x, y + S, S + 1) + square_sum(x, y + S - 1, S);
if (num[x][y] >= P)
num[x][y] -= P;
}
}
return num[sx][sy];
}
int main ()
{
scanf("%d", &TN);
for (TC = 1; TC <= TN; ++TC)
{
scanf("%d%d%d ", &N, &M, &S);
K = N + (S + 1) / 2;
for (int x = 0; x < N; ++x)
{
gets(tmp);
memcpy(board[x], tmp, N);
}
find_leaper(sx, sy);
int ans = solve();
printf("%d\n", ans);
}
}
In Java :
import java.util.Scanner;
public class Solution {
private static final int MOD_PRIME = 1000000007;
public static int addModulo(int x, int y) {
int s = x + y;
if (s >= MOD_PRIME)
s -= MOD_PRIME;
return s;
}
public static int subtractModulo(int x, int y) {
int d = x - y;
if (d < 0)
d += MOD_PRIME;
return d;
}
public static int query(int[][] sum, int x, int y, int s) {
int n = sum.length;
int rx = n / 2 + x - y;
int ry = x + y + 1;
int high_x = Math.min(rx + s, n - 1);
int low_x = Math.max(rx - s, 1);
int high_y = Math.min(ry + s, n - 1);
int low_y = Math.max(ry - s, 1);
int pos = addModulo(sum[high_x][high_y], sum[low_x - 1][low_y - 1]);
int neg = addModulo(sum[high_x][low_y - 1], sum[low_x - 1][high_y]);
return subtractModulo(pos, neg);
}
public static void createSumTable(int[][] sum, int[][] matrix) {
for (int i = 0; i < matrix.length; i++) {
int partialSum = 0;
for (int j = 0; j < matrix[i].length; j++) {
partialSum = addModulo(partialSum, matrix[i][j]);
sum[i][j] = partialSum;
if (i > 0)
sum[i][j] = addModulo(sum[i][j], sum[i - 1][j]);
}
}
}
public static final void solveProblem(Scanner in) {
String line = in.nextLine();
String tokens[] = line.split(" ");
int n = Integer.parseInt(tokens[0]);
int m = Integer.parseInt(tokens[1]);
int s = Integer.parseInt(tokens[2]);
int[][] rotated = new int[2 * n][2 * n];
int[][] sum = new int[2 * n][2 * n];
int[][] next = new int[2 * n][2 * n];
int lx = -1, ly = -1;
for (int i = 0; i < n; i++) {
line = in.nextLine();
for (int j = 0; j < line.length(); j++) {
char ch = line.charAt(j);
if (ch != 'P') {
rotated[n + i - j][i + j + 1] = 1;
if (ch == 'L') {
lx = i;
ly = j;
}
}
}
}
createSumTable(sum, rotated);
for (int k = 2; k <= m; k++) {
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++) {
int x = n + i - j;
int y = i + j + 1;
if (rotated[x][y] != 0)
next[x][y] = query(sum, i, j, s);
}
int[][] tmp = next;
next = rotated;
rotated = tmp;
createSumTable(sum, rotated);
}
System.out.println(query(sum, lx, ly, s));
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int numTests = Integer.parseInt(in.nextLine());
for (int i = 0; i < numTests; i++)
solveProblem(in);
in.close();
}
}
In C :
#include <stdio.h>
#define P 1000000007
long long ll,t,a[2][1000][1000],mm,mmm,p[1010][1010];
long long r[1010][1010],i,j,k,l,m,n;
char s[210][210];
/*
long long bb(long long ll, long long ii, long long jj)
{
if(ii<0 || ii>=n || jj<0 || jj>=n) return 0;
return a[ll][ii][jj];
}
*/
int main()
{
scanf("%lld",&t);
while(t--)
{
scanf("%lld %lld %lld\n",&n,&m,&k);
for(i=0;i<n;i++) scanf("%s\n",s[i]);
/*
for(i=0;i<n;i++)
for(j=0;j<n;j++)
for(l=0;l<0;l++) a[l][i][j]=0;
*/
for(i=0;i<1000;i++)
for(j=0;j<1000;j++)
a[0][i][j] = a[1][i][j] = 0;
for(i=0;i<n;i++)
for(j=0;j<n;j++)
if(s[i][j]=='L') a[0][i+500][j+500] = 1;
for(i=0;i<1000;i++)
for(j=0;j<1000;j++) p[i][j]= r[i][j] =0;
//for(l=0;l<m;l++)
l=0;
ll=1;
while(m--)
{
for(i=500;i<n+500;i++)
{
for(j=-k+500;j<n+500;j++)
{
p[i][j] = (p[i-1][j+1] + a[l][i][j] - a[l][i-1-k][j+1+k])%P;
}
}
for(i=n+500;i<n+k+500;i++)
{
for(j=500;j<n+500;j++)
{
p[i][j] = (p[i-1][j+1] + a[l][i][j] - a[l][i-1-k][j+1+k])%P;
}
}
for(i=n-1+500;i>=500-1;i--)
{
for(j=-k+500;j<=n+500;j++)
{
r[i][j] = (r[i+1][j+1] + a[l][i][j] - a[l][i+1+k][j+1+k])%P;
}
}
for(i=-2+500;i>=-k+500;i--)
{
for(j=500;j<=n+500;j++)
{
r[i][j] = (r[i+1][j+1] + a[l][i][j] - a[l][i+1+k][j+1+k])%P;
}
}
j = 500-1;
//jj = 500+j;
mmm=0;
for(i=-k+500;i<500;i++)
{
// ii = i+500;
mmm = (mmm + p[i+k][j])%P;
//+ r[i][j-k]
//- a[l][i+k][j]
//- p[i-1][j-k] - r[i-1-k][j]
//+ a[l][i-1-k][j])%P;
}
for(j=500;j<n+500;j++)
{
// jj = j+500;
i = 500-1;
// ii= 500+i;
mmm = (mmm + p[i+k][j] + r[i-k][j] - a[l][i][j+k]
- p[i][j-1-k] - r[i][j-1-k] + a[l][i][j-k-1])%P;
mm = mmm;
for(i=500;i<n+500;i++)
{
// ii = i+500;
mm = (mm + p[i+k][j] + r[i][j-k] - a[l][i+k][j]
- p[i-1][j-k] - r[i-1-k][j] + a[l][i-1-k][j])%P;
// printf("mm %lld %lld %lld\n",i,j,mm);
if(s[i-500][j-500]!='P') a[ll][i][j] = mm;
else a[ll][i][j] = 0;
}
}
l = (l+1)%2;
ll = (ll+1)%2;
/*
for(i=500;i<n+500;i++)
{
for(j=500;j<n+500;j++) printf("%lld ",a[l][i][j]);
printf("\n");
}
printf("--------------\n");
*/
}
mm = 0;
for(i=500;i<n+500;i++)
for(j=500;j<n+500;j++) mm = (mm + a[l][i][j])%P;
printf("%lld\n",(mm+P)%P);
//return 0;
}
return 0;
}
View More Similar Problems
Binary Search Tree : Insertion
You are given a pointer to the root of a binary search tree and values to be inserted into the tree. Insert the values into their appropriate position in the binary search tree and return the root of the updated binary tree. You just have to complete the function. Input Format You are given a function, Node * insert (Node * root ,int data) { } Constraints No. of nodes in the tree <
View Solution →Tree: Huffman Decoding
Huffman coding assigns variable length codewords to fixed length input characters based on their frequencies. More frequent characters are assigned shorter codewords and less frequent characters are assigned longer codewords. All edges along the path to a character contain a code digit. If they are on the left side of the tree, they will be a 0 (zero). If on the right, they'll be a 1 (one). Only t
View Solution →Binary Search Tree : Lowest Common Ancestor
You are given pointer to the root of the binary search tree and two values v1 and v2. You need to return the lowest common ancestor (LCA) of v1 and v2 in the binary search tree. In the diagram above, the lowest common ancestor of the nodes 4 and 6 is the node 3. Node 3 is the lowest node which has nodes and as descendants. Function Description Complete the function lca in the editor b
View Solution →Swap Nodes [Algo]
A binary tree is a tree which is characterized by one of the following properties: It can be empty (null). It contains a root node only. It contains a root node with a left subtree, a right subtree, or both. These subtrees are also binary trees. In-order traversal is performed as Traverse the left subtree. Visit root. Traverse the right subtree. For this in-order traversal, start from
View Solution →Kitty's Calculations on a Tree
Kitty has a tree, T , consisting of n nodes where each node is uniquely labeled from 1 to n . Her friend Alex gave her q sets, where each set contains k distinct nodes. Kitty needs to calculate the following expression on each set: where: { u ,v } denotes an unordered pair of nodes belonging to the set. dist(u , v) denotes the number of edges on the unique (shortest) path between nodes a
View Solution →Is This a Binary Search Tree?
For the purposes of this challenge, we define a binary tree to be a binary search tree with the following ordering requirements: The data value of every node in a node's left subtree is less than the data value of that node. The data value of every node in a node's right subtree is greater than the data value of that node. Given the root node of a binary tree, can you determine if it's also a
View Solution →