sgu 208 分类： sgu templates 2015-06-17 01:40 24人阅读 评论(0) 收藏

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感谢 Owaski 的帮助!

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设 G 是 p 个对象的一个置换群， S 为不同染色方案数

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Burnside引理：

令D(a) 表示在置换 a 下不变的元素的个数

S=∑D(i)|G|

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Polya定理：

如果用 m种颜色涂染 p个对象
令 C(a) 表示置换 a 中的循环节数

S=∑mC(i)|G|

时间复杂度为 O(|G|∗p)

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http://blog.csdn.net/qq_20118433/article/details/44220365

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在本题中，p=n∗m ,
|G|=n∗m∗4 | n=m , n∗m∗2 | n≠m
Polya 完全可以胜任，时间复杂度 O((n∗m)2)

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#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>

const int maxn = 23, maxm = 25, maxl = 100;
const int Base = 1e4, Blen = 4;

struct BigNum
{
    int num[maxl], l;

    void plus(int &hp,int &tp)
    {
        hp += tp/Base, tp %= Base;
    }
    void digit()
    {
        for(int i = 1; i <= l; i++) plus(num[i+1] ,num[i]);
        while(num[l+1]) ++l, plus(num[l+1], num[l]); 
    }
    void operator +=(const BigNum &add)
    {
        l = std::max(l, add.l);
        for(int i = 1; i <= l; i++)
            num[i] += add.num[i];

        digit();
    }
    void operator /=(const int x)
    {
        int nl = 0;
        for(int i = l; i > 0; i--)
        {
            num[i-1] += num[i]%x*Base, num[i] /= x;
            if(!nl && num[i]) nl = i;
        }
        l = nl;
    }
    void prt()
    {
        printf("%d",num[l]);
        for(int i = l-1; i > 0; i--)
            printf("%04d",num[i]);
    }
};

int n, m; 
bool square;int G;
int f[maxn*maxm];
BigNum pow2[maxn*maxm];
BigNum ans;

void PreWork()
{
    if(square = (n == m))
        G = (n*m)<<2;
    else
        G = (n*m)<<1;

    pow2[0].l = pow2[0].num[1] = 1;

    for(int i = 1; i <= n*m; i++)
        pow2[i] = pow2[i-1], pow2[i] += pow2[i-1];
}
void Slide_Down()
{
    static int tmp[maxn*maxm];

    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
            tmp[i%n*m+j] = f[(i-1)*m+j];

    for(int i = 1; i <= n*m; i++) f[i] = tmp[i];        
}
void Slide_Right()
{
    static int tmp[maxn*maxm];

    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
            tmp[(i-1)*m+j%m+1] = f[(i-1)*m+j];

    for(int i = 1; i <= n*m; i++) f[i] = tmp[i];            
}
void Rotate()
{
    static int tmp[maxn*maxm];

    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
            tmp[(j-1)*n+n-i+1] = f[(i-1)*m+j];

    for(int i = 1; i <= n*m; i++) f[i] = tmp[i];    
    std::swap(n, m);            
}
int calc()
{
    static bool hash[maxn*maxm];
    memset(hash,false,sizeof(hash));

    int cnt = 0;

    for(int i = 1; i <= n*m; i++)
        for(int j = i; !hash[j]; j = f[j])
        {
            if(i == j) cnt++;
            hash[j] = true;
        }

    return cnt; 
}
void Polya()
{
    for(int i = 1; i <= n*m; i++) f[i] = i;

    for(int i = 1; i <= n; i++, Slide_Down())
        for(int j = 1; j <= m; j++, Slide_Right())
        {
            ans += pow2[calc()]; 
            Rotate();
            if(square) ans += pow2[calc()];
            Rotate();
            ans += pow2[calc()];
            Rotate();
            if(square) ans += pow2[calc()];
            Rotate();
        }

    ans /= G;   
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("sgu208.in","r",stdin);
    freopen("sgu208.out","w",stdout);
#endif

    std::cin >> n >> m; 

    PreWork(), Polya();

    ans.prt();

#ifndef ONLINE_JUDGE
    fclose(stdin);
    fclose(stdout);
#endif
}

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