# Training Little Cats (POJ 3735)

If we regard every operation on cats as matrix multiplication, all operations in a round can be represented as a single matrix. To repeat m round is to calculate the nth power of that matrix.

The principle is simple but there are some problems that need to be overcomed. At first, the result might be very large and exceed the range of int type so we have to use long. Secondly, we have to improve the time efficiency of calculate matrix multiplication as the matrix may be very large. After observation we find that the matrix of a single round is a sparse matrix as every row at most has two non-zero elements. And after each mutiplication, this property won't change a lot. So we should skip the zeros in the matrix when multiplying to reduce the time complexity from to be very near to . Only after these two points is taken care of, we can pass the test case on POJ.

import java.io.*;
import java.util.*;

public class TrainingLittleCats {
private static long[][] multiple(long[][] A, long[][] B) {
long[][] C = new long[A.length][B.length];
for (int i = 0; i < A.length; i++) {
for (int k = 0; k < B.length; k++) {
if (A[i][k] == 0) continue; // IMPORTANT: optimize time cost!
for (int j = 0; j < B.length; j++) {
C[i][j] = (C[i][j] + A[i][k] * B[k][j]);
}
}
}
return C;
}

private static long[][] pow(long[][] A, long n) {
long[][] I = new long[A.length][A.length];
for (int i = 0; i < A.length; i++) {
I[i][i] = 1;
}

while (n > 0) {
if ((n & 1) == 1) I = multiple(I, A);
A = multiple(A, A);
n >>= 1;
}
return I;
}

private static long[][] makeIdentity(int n) {
long[][] I = new long[n][n];
for (int i = 0; i < n; i++) {
I[i][i] = 1;
}
return I;
}

private static void give(long[][] A, int i) {
A[i]++;
}

private static void eat(long[][] A, int i) {
for (int j = 0; j < A.length; j++) {
A[i][j] = 0;
}
}

private static void swap(long[][] A, int i, int j) {
long[] temp = A[i];
A[i] = A[j];
A[j] = temp;
}

public static void main(String[] args) {
Scanner in = new Scanner(System.in);

boolean flag = false;

while(in.hasNextInt()) {
int n = in.nextInt() + 1, m = in.nextInt(), k = in.nextInt();

if (n == 1) return;

if (flag) System.out.println();
else flag = true;

long[][] A = makeIdentity(n);

for (int i = 0; i < k; i++) {
String op = in.next();
if (op.equals("g")) {
give(A, in.nextInt());
} else if (op.equals("e")) {
eat(A, in.nextInt());
} else {
swap(A, in.nextInt(), in.nextInt());
}
}

A = pow(A, m);
System.out.printf("%d", A);
for (int i = 2; i < n; i++) {
System.out.printf(" %d", A[i]);
}
}
}
}