这道题就是连通性状态压缩DP，复习了一下。
#include <cstdio>
#include <iostream>
#include <cstring>
#include <map>
using namespace std;
long long dp[11][11][(1<<11)+1][2], n, m;
long long dfs(int x, int y, int md, int right){long long &ret = dp[x][y][md][right];if(ret != -1) return ret;if(x >= n){if(!md) return ret = 1;else return ret = 0;}if(y >= m){if(right) return ret = 0;else return ret = dfs(x+1, 0, md, 0);}ret = 0;int now_Pos = 1 << y;if(right){if(md & now_Pos) return ret = 0;return ret = dfs(x, y+1, md, 0);}if(md & now_Pos){int n_md = md ^ now_Pos;return ret = dfs(x, y+1, n_md, right);}int n_md = md | now_Pos;return ret =( y+1 < m ? dfs(x, y+1, md, 1) : 0 )+ dfs(x, y+1, n_md, 0);
}
int main(){while(scanf("%d %d", &n, &m)!=EOF){if(n == 1 || m == 1){if((n % 2)^(m % 2))cout<<1<<endl;else cout<<0<<endl;continue;}memset(dp, -1, sizeof dp);cout<<dfs(0, 0, 0, 0)<<endl;}
}这道题如果写递归版本的要TLE不知为何，话说我还没有A
