POJ 3254 状压dp

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#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <vector>using namespace std;const int maxn = 1 << 12 + 10;const int MOD = 1E8;int dp[15][maxn], mp[15][15], n, m;std::vector<int> vec[15];int fun(int x){int s = 0;for (int i = 1; i <= m; i++)s += (!mp[x][i]) * (1 << (m - i));return s;}int main(int argc, char const *argv[]){while (~scanf("%d%d", &n, &m) && n + m){memset(dp, 0, sizeof(dp));memset(vec, 0, sizeof(vec));for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++)scanf("%d", &mp[i][j]);vec[0].push_back(0);for (int i = 0; i < (1 << m); i++)dp[0][i] = 1;for (int i = 1; i <= n; i++){int tmp = fun(i);for (int j = 0; j < (1 << m); j++)if (!((j & (j >> 1)) || (j & tmp)))vec[i].push_back(j);for (int j = 0; j < vec[i].size(); j++){int u = vec[i][j];for (int k = 0; k < vec[i - 1].size(); k++){int v = vec[i - 1][k];if (v & u) continue;dp[i][u] = (dp[i][u] + dp[i - 1][v]) % MOD;}}}int ans = 0;for (int i = 0; i < (1 << m); i++)ans = (ans + dp[n][i]) % MOD;printf("%d\n", ans);}return 0;}

dp[i][j] 就表示第i行状态为j时的方案数。

判断第i行是不是有两块相邻的土地同时都有牛，假设当前状态为X,那么只需要判断X&(X>>1)的结果，如果是0，说明没有相邻的，否则就说明有相邻的。

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