题目传送门
1 /*
2 状态压缩DP：先处理硬性条件即不能种植的，然后处理左右不相邻的，
3 接着就是相邻两行查询所有可行的种数并累加
4
5 写错一个地方差错N久:)
6 详细解释：http://www.tuicool.com/articles/JVzMVj
7 */
8 #include <cstdio>
9 #include <iostream>
10 #include <algorithm>
11 #include <cmath>
12 #include <cstring>
13 #include <string>
14 #include <map>
15 #include <set>
16 using namespace std;
17
18 const int MAXN = 15;
19 const int INF = 0x3f3f3f3f;
20 const int MOD = 1e8;
21 int dp[MAXN][1 << MAXN];
22 int st[1 << MAXN];
23 int rem[1 << MAXN];
24
25 int solve(int m)
26 {
27 int cnt = 0;
28 int tot = 1 << m;
29 for (int i=0; i<tot; ++i)
30 {
31 if ((i & (i << 1)) == 0)
32 {
33 st[++cnt] = i;
34 }
35 }
36
37 return cnt;
38 }
39
40 void work(int n, int m, int cnt)
41 {
42 memset (dp, 0, sizeof (dp));
43 for (int i=1; i<=cnt; ++i)
44 {
45 if ((st[i] & rem[1]) == 0)
46 {
47 dp[1][i] = 1;
48 }
49 }
50
51 for (int i=2; i<=n; ++i)
52 {
53 for (int k=1; k<=cnt; ++k)
54 {
55 if ((st[k] & rem[i]) == 0)
56 {
57 for (int j=1; j<=cnt; ++j)
58 {
59 if (((st[j] & rem[i-1]) == 0) && ((st[j] & st[k]) == 0))
60 dp[i][k] = (dp[i][k] + dp[i-1][j]) % MOD;
61 }
62 }
63 }
64 }
65
66 int ans = 0;
67 for (int i=1; i<=cnt; ++i)
68 {
69 ans = (ans + dp[n][i]) % MOD;
70 }
71
72 printf ("%d
", ans);
73 }
74
75 int main(void) //POJ 3254 Corn Fields
76 {
77 #ifndef ONLINE_JUDGE
78 freopen ("F.in", "r", stdin);
79 #endif
80
81 int n, m;
82 while (~scanf ("%d%d", &n, &m))
83 {
84 int cnt = solve (m);
85
86 memset (rem, 0, sizeof (rem));
87 for (int i=1; i<=n; ++i)
88 {
89 int x;
90 for (int j=1; j<=m; ++j) //看错n和m
91 {
92 scanf ("%d", &x);
93 if (x == 0)
94 {
95 rem[i] = rem[i] | (1 << (m-j));
96 }
97 }
98 }
99
100 work (n, m, cnt);
101 }
102
103 return 0;
104 }