其实这题还能用状压DP解决,可是时间达到2000ms只能过掉POJ2411.状压DP解法详见状压DP解POJ2411
贴上POJ2411AC代码 : 2000ms 时间复杂度h*w*(2^w)*(2^w)
#include <cstdio> #include <cmath> #include <algorithm> #include <cstring> #include <utility> #include <string> #include <iostream> #include <map> #include <set> #include <vector> #include <queue> #include <stack> using namespace std; #pragma comment(linker, "/STACK:1024000000,1024000000") #define eps 1e-10 #define inf 0x3f3f3f3f #define PI pair<int, int> typedef long long LL; const int maxn = 1 << 11; LL dp[12][maxn]; int w, h; bool TestFirstLine(int state) { for(int i = 0; i < w;) { int x = 1 << i; if((x & state) && i+1 < w && ((x << 1) & state)) { i += 2; } else if(!(x & state)) ++i; else return false; } return true; } bool is_ok(int state1, int state2) { for(int i = 0; i < w;) { int x = 1 << i; if(!(x & state1)) { if(!(x & state2)) return false; ++i; } else { int y = x << 1; if(!(x & state2)) ++i; else if(x & state2) { if(i == w-1 || !(y & state1) || !(y & state2)) return false; i += 2; } } } return true; } LL solve() { if(h < w) swap(w, h); int r = 1 << w; memset(dp, 0, sizeof(dp)); //边界 for(int i = 0; i < r; ++i) if(TestFirstLine(i)) dp[0][i] = 1; for(int i = 1; i < h; ++i) for(int j = 0; j < r; ++j) for(int k = 0; k < r; ++k) { if(is_ok(j, k)) dp[i][j] += dp[i-1][k]; } return dp[h-1][r-1]; } int main() { while(scanf("%d%d", &w, &h) == 2 && w && h) { printf("%lld\n", solve()); } return 0; }
对于uva11270这种状压dp会超时,使用轮廓线DP可将复杂度降低到w*h*(2^w),详细解法见训练指南P384
AC代码
#include <cstdio> #include <cmath> #include <algorithm> #include <cstring> #include <utility> #include <string> #include <iostream> #include <map> #include <set> #include <vector> #include <queue> #include <stack> using namespace std; #pragma comment(linker, "/STACK:1024000000,1024000000") #define eps 1e-10 #define inf 0x3f3f3f3f #define PI pair<int, int> typedef long long LL; const int maxn = 10 + 1; LL dp[2][1<<maxn]; int n, m; void update(int f, int a, int b) { if(b & (1<<m)) dp[f][b^(1<<m)] += dp[1-f][a]; } LL solve() { if(n < m) swap(n, m); memset(dp, 0, sizeof(dp)); int f = 0; dp[0][(1<<m)-1] = 1; //边界 for(int i = 0; i < n; ++i) for(int j = 0; j < m; ++j) { f ^= 1; memset(dp[f], 0, sizeof(dp[f])); for(int k = 0; k < (1 << m); ++k) { //不放 update(f, k, k<<1); //横着放 if(j && !(k&1)) update(f, k, (k<<1)^3); //竖着放 if(i && !(k&(1<<m-1))) update(f, k, (k<<1)^(1<<m)^1); } } return dp[f][(1<<m)-1]; } int main() { while(scanf("%d%d", &n, &m) == 2) { printf("%lld\n", solve()); } return 0; }
如有不当之处欢迎指出!