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  • BZOJ4000 TJOI2015棋盘(状压dp+矩阵快速幂)

      显然每一行棋子的某种放法是否合法只与上一行有关,状压起来即可。然后n稍微有点大,矩阵快速幂即可。

    #include<iostream> 
    #include<cstdio>
    #include<cmath>
    #include<cstdlib>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    int read()
    {
        int x=0,f=1;char c=getchar();
        while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
        while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
        return x*f;
    }
    #define N 1000010
    #define ul unsigned int
    #define S 64
    int n,m,p,k;
    bool flag[3][6];
    struct matrix
    {
        int n;ul a[S][S];
        matrix operator *(const matrix&b) const
        {
            matrix c;c.n=n;memset(c.a,0,sizeof(c.a));
            for (int i=0;i<n;i++)
                for (int j=0;j<S;j++)
                    for (int k=0;k<S;k++)
                    c.a[i][j]+=a[i][k]*b.a[k][j];
            return c;
        }
    }f,a;
    int main()
    {
    #ifndef ONLINE_JUDGE
        freopen("bzoj4000.in","r",stdin);
        freopen("bzoj4000.out","w",stdout);
        const char LL[]="%I64d
    ";
    #else
        const char LL[]="%lld
    ";
    #endif
        n=read(),m=read(),p=read(),k=read();
        for (int i=0;i<3;i++)
            for (int j=0;j<p;j++)
            flag[i][j]=read();
        flag[1][k]=0;
        f.n=1;f.a[0][0]=1;
        a.n=1<<m;
        for (int i=0;i<(1<<m);i++)
            for (int j=0;j<(1<<m);j++)
            {
                a.a[i][j]=1;
                for (int v=0;v<m;v++)
                if (i&(1<<v))
                    for (int x=max(0,v-k);x<min(m,v+p-k);x++)
                    {    
                        if (flag[1][x-(v-k)]&&(i&(1<<x))) a.a[i][j]=0;
                        if (flag[2][x-(v-k)]&&(j&(1<<x))) a.a[i][j]=0;
                    }
                for (int v=0;v<m;v++)
                if (j&(1<<v))
                    for (int x=max(0,v-k);x<min(m,v+p-k);x++)
                    {    
                        if (flag[1][x-(v-k)]&&(j&(1<<x))) a.a[i][j]=0;
                        if (flag[0][x-(v-k)]&&(i&(1<<x))) a.a[i][j]=0;
                    }
            }
        for (;n;n>>=1,a=a*a) if (n&1) f=f*a;
        ul ans=0;
        for (int i=0;i<(1<<m);i++) ans+=f.a[0][i];
        cout<<ans;
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Gloid/p/9823478.html
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