poj2411 Mondriaan's Dream(状压dp)
2017-10-12 21:46
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用1*2的小木块覆盖n*m的格子,问方案数。n,m<=10.状压dp,用二进制数表示一行的状态,逐行转移。横放用11表示,竖放用01表示。a[s]存的是上一行状态为s,能转移到的所有状态。初始值f[0][tot]=1,表示第0行全为1的方案有1个。最后答案就是f
[tot]。复杂度不会算啊。。。O(T*2^n*n*2^n)???
[tot]。复杂度不会算啊。。。O(T*2^n*n*2^n)???
#include <cstdio> #include <cstring> #include <vector> using namespace std; #define ll long long #define N 12 #define inf 0x3f3f3f3f inline int read(){ int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); return x*f; } int tot; int n,m,bin ; ll f [1<<11]; vector<int>a[1<<11]; void dfs(int s,int pos,int ss){ if(pos>m){a[s].push_back(ss);return;} //横放1*2 if(s&bin[pos-1]&&s&bin[pos]) dfs(s,pos+2,ss|bin[pos-1]|bin[pos]); //竖放2*1 if(s&bin[pos-1]) dfs(s,pos+1,ss); else dfs(s,pos+1,ss|bin[pos-1]); } int main(){ // freopen("a.in","r",stdin); bin[0]=1; for(int i=1;i<=11;++i) bin[i]=bin[i-1]<<1; while(1){ n=read();m=read();if(!n) break;tot=bin[m]-1; for(int i=0;i<=tot;++i) a[i].clear(),dfs(i,1,0); memset(f,0,sizeof(f));f[0][bin[m]-1]=1;ll ans=0; for(int i=0;i<=n-1;++i) for(int s=0;s<=tot;++s){ if(!f[i][s]) continue; for(int j=0;j<a[s].size();++j) f[i+1][a[s][j]]+=f[i][s]; } printf("%lld\n",f [tot]); }return 0; }
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