德玛西亚万岁

https://ac.nowcoder.com/acm/problem/15034
个人体会:题目卡错，多写半年(忘记取模和多组数据QAQ)
思路:通过题目的描述，又是给棋盘数据范围又20不到，肯定是状压dp
状态表示:
dp[i][j]:前i行中，且第i行的放置状态为j的所有方法的cnt
状态计算:
dp[i][j] += dp[i - 1][k](前提相邻行的士兵不能相邻，即j & k == 0)
状态dp的一般流程:设置好状态表示和计算后，先预处理出的合法状态，本题中对于任意一行，其合法条件为
1.棋盘为0的位置不放置士兵，即(g[i] & j == g[i]
2.同一排士兵不能相邻 即((j >> 1) & j) == 0 && ((j << 1) & j))

AC代码

#include <bits/stdc++.h>
using namespace std;
#define IO std::ios::sync_with_stdio(false); cin.tie(0)
#define ll long long
#define ull unsigned long long
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
#define rep(i, l, r) for (int i = l; i <= r; ++i)
#define per(i, l, r) for (int i = l; i >= r; --i)
#define mset(s, _) memset(s, _, sizeof(s))
#define mcpy(s, _) memcpy(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define vi vector<int>
#define vpii vector<pii> 
#define mp(a, b) make_pair(a, b)
#define pll pair <ll, ll>
#define fir first
#define sec second
#define inf 0x3f3f3f3f
inline int lowbit(int x) {return x & -x;} 
template< typename T > inline void get_min(T &x, T y) {if(y < x) x = y;}
template< typename T > inline void get_max(T &x, T y) {if(x < y) x = y;}
inline int read() {
  int x = 0, f = 0; char ch = getchar();
  while (!isdigit(ch)) f |= ch == '-', ch = getchar();
  while (isdigit(ch)) x = 10 * x + ch - '0', ch = getchar();
  return f ? -x : x;
}
template<typename T> inline void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
template<typename T> inline void print(T x, char let) {
  print(x), putchar(let);
}

const int N = 15, M = 1e7 + 10, mod = 100000000;
int n, m;
int g[N];
ll dp[N][1 << N];
vi leg[N];

void init() {
    mset(dp, 0);
    mset(g, 0);
    rep(i, 0, n + 1) leg[i].clear();
}
int main() {
    while(cin >> n >> m) {
        init();
        for(int i = 1; i <= n; i ++ ) {
            for(int j = 1; j <= m; j ++ ) {
                int t; cin >> t;  g[i] += (t << (j - 1));
            }
        } 

        for(int i = 1; i <= n + 1; i ++ ) {
            for(int j = 0; j < (1 << m); j ++ ) {
                if((g[i] | j) == g[i] && ((j << 1) & j) == 0 && ((j >> 1) & j) == 0) {
                    leg[i].pb(j);
                    if(i == 1) dp[1][j] = 1;
                }
            }
        }

        for(int i = 2; i <= n + 1; i ++ ) {
            for(auto j : leg[i]) {
                for(auto k : leg[i - 1]) {
                    if((j & k) == 0) {
                        dp[i][j] = (dp[i][j] + dp[i - 1][k]) % mod;
                    }
                }
            }
        }

        cout << dp[n + 1][0] << endl;
    }

    return 0;
}