没什么可说的,入门级状压DP。直接撸掉
#include <iostream> #include <cstring> #include <cstdlib> #include <cstdio> #include <algorithm> #include <cmath> #include <vector> #include <map> #include <queue> #include <stack> #include <set> #define LL long long #define FOR(i, x, y) for(int i=x;i<=y;i++) using namespace std; const int maxn = 5000; const int MOD = 100000000; int N, M, K; int C[20], state[maxn]; int dp[20][maxn]; bool judge(int x) { if(x & (x << 1)) return false; return true; } void init() { memset(dp, 0, sizeof(dp)); memset(state, 0, sizeof(state)); memset(C, 0, sizeof(C)); int r = (1 << M) - 1; K = 0; FOR(i, 0, r) if(judge(i)) state[++K] = i; } int main() { while(scanf("%d %d", &N, &M)!=EOF) { init(); int X; FOR(i, 1, N) FOR(j, 1, M) { scanf("%d", &X); if(X == 0) C[i] = C[i] | (1 << (M - j)); } FOR(i, 1, K) { if(!(state[i] & C[1])) dp[1][i] = 1; } FOR(i, 2, N) FOR(j, 1, K) { if(C[i-1] & state[j]) continue; FOR(l, 1, K) { if((C[i] & state[l]) || (state[l] & state[j])) continue; dp[i][l] = (dp[i-1][j] + dp[i][l]) % MOD; } } int ans = 0; FOR(i, 1, K) ans = (ans + dp[N][i]) % MOD; printf("%d ", ans); } return 0; }