[HDU 5731]Solid Dominoes Tilings：状压DP+容斥原理

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这题是POJ 2411的升级版。dp[i][j]表示长为i宽为j的方案数，打个表预处理一下。然后枚举列的切割方案，因为一共有m-1条竖线，因此有2^(m-1)种情况。cnt[i]表示按当前的列分割方案，宽为i的矩形的方案数，f[i]表示在当前列分割情况下，前i行的情况数，f[i]=cnt[i]-sigma{f[j]*cnt[i-j]}( 0 < j < i )

/*User:SmallLanguage:C++Problem No.:5731*/#include<bits/stdc++.h>#define ll long long#define inf 999999999using namespace std;const ll mod=1e9+7;ll dp[17][17]={0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,0,0,3,0,11,0,41,0,153,0,571,0,2131,0,7953,0,29681,0,1,5,11,36,95,281,781,2245,6336,18061,51205,145601,413351,1174500,3335651,9475901,0,0,8,0,95,0,1183,0,14824,0,185921,0,2332097,0,29253160,0,366944287,0,1,13,41,281,1183,6728,31529,167089,817991,4213133,21001799,106912793,536948224,720246619,704300462,289288426,0,0,21,0,781,0,31529,0,1292697,0,53175517,0,188978103,0,124166811,0,708175999,0,1,34,153,2245,14824,167089,1292697,12988816,108435745,31151234,940739768,741005255,164248716,498190405,200052235,282756494,0,0,55,0,6336,0,817991,0,108435745,0,479521663,0,528655152,0,764896039,0,416579196,0,1,89,571,18061,185921,4213133,53175517,31151234,479521663,584044562,472546535,732130620,186229290,274787842,732073997,320338127,0,0,144,0,51205,0,21001799,0,940739768,0,472546535,0,177126748,0,513673802,0,881924366,0,1,233,2131,145601,2332097,106912793,188978103,741005255,528655152,732130620,177126748,150536661,389322891,371114062,65334618,119004311,0,0,377,0,413351,0,536948224,0,164248716,0,186229290,0,389322891,0,351258337,0,144590622,0,1,610,7953,1174500,29253160,720246619,124166811,498190405,764896039,274787842,513673802,371114062,351258337,722065660,236847118,451896972,0,0,987,0,3335651,0,704300462,0,200052235,0,732073997,0,65334618,0,236847118,0,974417347,0,1,1597,29681,9475901,366944287,289288426,708175999,282756494,416579196,320338127,881924366,119004311,144590622,451896972,974417347,378503901};ll f[17];int b[20],cnt,siz,n,m;ll solve(int n,int m){    ll res=0;    for(int i=0;i<(1<<m-1);i++){        cnt=siz=0;        for(int j=0;j<m-1;j++){            siz++;            if(i>>j&1){                b[++cnt]=siz;                siz=0;            }        }        b[++cnt]=++siz;        for(int j=1;j<=n;j++){            for(int k=0;k<j;k++){                ll tmp=1;                for(int t=1;t<=cnt;t++)                    tmp=tmp*dp[b[t]][j-k]%mod;                if(!k) f[j]=tmp;                else f[j]=((f[j]-f[k]*tmp%mod)%mod+mod)%mod;            }        }        if(!(cnt&1)) res=((res-f[n])%mod+mod)%mod;        else res=(res+f[n])%mod;    }    return res;}int main(){    freopen("data.in","r",stdin);//    ios::sync_with_stdio(false);    while(cin>>n>>m) cout<<solve(n,m)<<endl;     return 0;}