UVA

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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;}

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