POJ 3254. Corn Fields
看来状压dp入门题都是这个难度的。
令dp[i][j]表示到了第 i 行,状态为 j 时的方案数。
然后枚举这一行和上一行的状态,则dp[i][j] += dp[i - 1][k]。
预处理合法的种植状态。
还有,判断种植状态符不符合草地状态只能一位一位比较,因为位运算不能辨别两个数的二进制,只能得出他们的运算结果。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 #include<algorithm> 5 #include<cstring> 6 #include<cstdlib> 7 #include<cctype> 8 #include<vector> 9 #include<stack> 10 #include<queue> 11 using namespace std; 12 #define enter puts("") 13 #define space putchar(' ') 14 #define Mem(a) memset(a, 0, sizeof(a)) 15 typedef long long ll; 16 typedef double db; 17 const int INF = 0x3f3f3f3f; 18 const db eps = 1e-8; 19 const int mod = 1e8; 20 const int max_sta = 400; 21 inline ll read() 22 { 23 ll ans = 0; 24 char ch = getchar(), last = ' '; 25 while(!isdigit(ch)) {last = ch; ch = getchar();} 26 while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();} 27 if(last == '-') ans = -ans; 28 return ans; 29 } 30 inline void write(ll x) 31 { 32 if(x < 0) x = -x, putchar('-'); 33 if(x >= 10) write(x / 10); 34 putchar(x % 10 + '0'); 35 } 36 37 int n, m; 38 int Map[15], s[max_sta], sum[max_sta], cnt = 0; 39 ll dp[15][max_sta]; 40 41 bool judge(int a, int b) 42 { 43 for(int i = 0; i < m; ++i) 44 if(!(a & (1 << i)) && (b & (1 << i))) return 0; 45 return 1; 46 } 47 48 int getsum(int x) 49 { 50 int ret = 0; 51 for(; x; x >>= 1) ret += x & 1; 52 return ret; 53 } 54 void init() 55 { 56 for(int i = 0; i < (1 << m); ++i) 57 if(!(i & (i << 1)) && !(i & (i >> 1))) 58 { 59 s[++cnt] = i; 60 if(judge(Map[1], i)) dp[1][cnt]++; 61 } 62 } 63 64 int main() 65 { 66 n = read(); m = read(); 67 for(int i = 1; i <= n; ++i) 68 for(int j = 0; j < m; ++j) 69 { 70 int x = read(); 71 Map[i] |= x << j; 72 } 73 init(); 74 for(int i = 2; i <= n; ++i) 75 for(int j = 1; j <= cnt; ++j) if(judge(Map[i], s[j])) 76 for(int k = 1; k <= cnt; ++k) if(judge(Map[i - 1], s[k]) && !(s[k] & s[j])) 77 (dp[i][j] += dp[i - 1][k]) %= mod; 78 ll ans = 0; 79 for(int i = 1; i <= cnt; ++i) (ans += dp[n][i]) %= mod; 80 write(ans); enter; 81 return 0; 82 }