# Corn Fields (POJ 3254)

A state compressed dynamic programming problem. We can use a binary integer to represent the state of a row in the fields and keep the count of ways under each state. Then for each row, we iterate new state and count based on previous row's count and state. At first we added a mock row before the first row with only the empty state has count 1 to simplify the implementing.

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

public class CornFields {
private static long MOD = 100000000;
private static void solve(int N, int M, int[] rows) {
if (N == 0 || M == 0) {
System.out.println(1);
return;
}
int W = (1 << M) - 1;
long[] last = new long[1 << M];
long[] cur = new long[1 << M];

last[0] = 1;

for (int i = 0; i < N; i++) {
for (int j = 0; j <= W; j++) {
if (last[j] == 0) continue;

for (int k = 0; k <= W; k++) {
if ((k & rows[i]) == 0 && (j & k) == 0 && (k & (k << 1)) == 0) {
cur[k] = (cur[k] + last[j]) % MOD;
}
}
}

long[] tmp = last;
last = cur;
cur = tmp;
Arrays.fill(cur, 0);
}

long res = 0;
for (int i = 0; i < last.length; i++) {
res = (res + last[i]) % MOD;
}
System.out.println(res);
}

public static void main(String[] args) {
Scanner in = new Scanner(System.in);
while (in.hasNextInt()) {
int N = in.nextInt(); int M = in.nextInt();
int[] rows = new int[N];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
if (in.nextInt() == 0) rows[i] |= (1 << j);
}
}
solve(N, M, rows);
}
}
}