POJ 3254 Corn Fields (状压dp)

 1 //#pragma comment(linker, "/STACK:102400000, 102400000")
 2 #include <algorithm>
 3 #include <iostream>
 4 #include <cstdlib>
 5 #include <cstring>
 6 #include <cstdio>
 7 #include <vector>
 8 #include <cmath>
 9 #include <ctime>
10 #include <list>
11 #include <set>
12 #include <map>
13 using namespace std;
14 int dp[13][1 << 12], mod = 1e8, state[13];
15 
16 int main()
17 {
18     int n, m;
19     while(~scanf("%d %d", &n, &m)) {
20         int num;
21         for(int i = 1; i <= n; ++i) {
22             state[i] = 0;
23             for(int j = 1; j <= m; ++j) {
24                 scanf("%d", &num);
25                 state[i] = (state[i] << 1) + num; // 初始可行状态
26             }
27         }
28         memset(dp, 0, sizeof(dp));
29         for(int i = 0; i < (1 << m); ++i) { // 初始化第一行
30             if((i & (i << 1)) == 0 && (i & state[1]) == i) { // 相邻没菜 且在可行状态中
31                 dp[1][i] = 1;
32             }
33         }
34         for(int i = 2; i <= n; ++i) {
35             for(int j = 0; j < (1 << m); ++j) {
36                 if((j & (j << 1)) == 0 && (j & state[i]) == j) {
37                     for(int k = 0; k < (1 << m); ++k) {
38                         if((j & k) == 0 && (k & state[i - 1]) == k) { //上下行菜不相邻 且上行可行
39                             dp[i][j] = (dp[i][j] + dp[i - 1][k]) % mod;
40                         }
41                     }
42                 }
43             }
44         }
45         int ans = 0;
46         for(int i = 0; i < (1 << m); ++i) {
47             ans = (dp[n][i] + ans) % mod;
48         }
49         printf("%d\n", ans);
50     }
51     return 0;
52 }