今日SGU 5.15

最近事情好多，数据库作业，没天要学2个小时java，所以更新的sgu就比较少了

SGU 131

题意：给你两种小块一种，1*1，一种2*2-1*1，问你填满一个m*n的矩形有多少钟方法，n和m小于等于9,

收获：状态压缩，每一行都最多由上一行转移过来，因为上一行，那么最多有7情况，详情看代码

#include<bits/stdc++.h>
#define de(x) cout<<#x<<"="<<x<<endl;
#define dd(x) cout<<#x<<"="<<x<<" ";
#define rep(i,a,b) for(int i=a;i<(b);++i)
#define repd(i,a,b) for(int i=a;i>=(b);--i)
#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)
#define ll long long
#define mt(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
#define pii pair<int,int>
#define pdd pair<double,double>
#define pdi pair<double,int>
#define mp(u,v) make_pair(u,v)
#define sz(a) (int)a.size()
#define ull unsigned long long
#define ll long long
#define pb push_back
#define PI acos(-1.0)
#define qc std::ios::sync_with_stdio(false)
#define db double
#define all(a) a.begin(),a.end()
const int mod = 1e9+7;
const int maxn = 11;
const double eps = 1e-6;
using namespace std;
bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }
bool ls(const db &a, const db &b) { return a + eps < b; }
bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }
ll gcd(ll a,ll b) { return a==0?b:gcd(b%a,a); };
ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }
ll kpow(ll a,ll b) {ll res=1;a%=mod; if(b<0) return 1; for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll read(){
    ll 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;
}
//inv[1]=1;
//for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
int n,m;
ll dp[maxn][1<<maxn];
void dfs(int i,int b1,int b2,int s1,int s2,int k){
    //b1,b2分别代表上行的前一列对这列的影响，和这行前一列对这列的影响
    //s1表示这行达到s2状态上一行需要满足的状态
    if(k==m){
        if(!b1&&!b2) dp[i][s2]+=dp[i-1][s1];
        return;
    }
    if(!b1&&!b2){
        dfs(i,0,0,s1<<1,s2<<1|1,k+1);
        dfs(i,0,1,s1<<1,s2<<1|1,k+1);
        dfs(i,1,0,s1<<1,s2<<1|1,k+1);
    }
    if(!b1) dfs(i,1,1,s1<<1,s2<<1|b2,k+1);
    if(!b2){
        dfs(i,0,1,s1<<1|(1-b1),s2<<1|1,k+1);
        dfs(i,1,1,s1<<1|(1-b1),s2<<1|1,k+1);
    }
    dfs(i,0,0,s1<<1|(1-b1),s2<<1|b2,k+1);
}
int main(){
    scanf("%d%d",&n,&m);
    dp[0][(1<<m)-1] = 1;
    rep(i,1,n+1) dfs(i,0,0,0,0,0);
    printf("%lld\n",dp[n][(1<<m)-1]);
    return 0;
}

原文地址：https://www.cnblogs.com/chinacwj/p/9042784.html