题意：

思路：

#include <cstdio>
#include <cstring>
#include <vector>
#include <iostream>
using namespace std;
const int N = 15;
const int mod = 100000000;
int s[N];//每行的状态
int dp[N][4100];//dp(i,j)排到第i行，且第i行的状态为j的合法安排方案
int n,m;
vector<int> v;
//筛选所有合法状态 
void init(){
    v.clear();
    for(int i = 0;i < (1<<m);i++){
        if(i & i<<1) continue;
        v.push_back(i); 
    }
} 
int main(){
    while(~scanf("%d%d",&n,&m)){
        for(int i = 1;i <= n;i++){
            int state = 0;
            for(int j = 0,x;j < m;j++){
                scanf("%d",&x);
                state <<= 1;
                if(x) state |= 1; 
            }
            s[i] = state;
        }
        init();
        memset(dp,0,sizeof(dp));
        dp[0][0] = 1;
        //s1上一行的状态，s2这一行的状态 
        int s1,s2; 
        for(int i = 1;i <= n+1;i++){
            for(int j = 0;j < v.size();j++){
                s1 = v[j];
                if((s1&s[i]) != s1) continue;
                for(int k = 0;k < v.size();k++){
                    s2 = v[k];
                    if(s1&s2) continue;
                    dp[i][s1] = (dp[i][s1] + dp[i-1][s2]) % mod;
                }
            }
        }
        printf("%d\n",dp[n+1][0]);
    }
    return 0;
}