bzoj 1004: [HNOI2008]Cards

    这也是一道polya定理的题，只不过在求循环节数的时候由于有使用个数限制，所以不能直接快速幂，而是用DP求出每个置换的循环节。DP很简单，近乎于暴力=_=

    上代码：

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <iostream>

#include <algorithm>

#define N 100

#define M 25

using namespace std;

 

int n, m, a[N];

int sb, sr, sg;

int yu;

int f[M][M][M];

 

int mi(int a, int b)

{

    int ans = 1, zan = a;

    while (b)

    {

        if (b & 1)

        {

            ans *= zan;

            if (ans > yu) ans %= yu;

        }

        b >>= 1;

        zan *= zan;

        if (zan > yu) zan %= yu;

    }

    return ans;

}

 

int cal()

{

    memset(f, 0, sizeof(f));

    int cir[N] = {0}, cirnum = 0;

    bool vis[N] = {0};

    for (int i = 1; i <= n; ++i)

        if (!vis[i])

        {

            cirnum ++; cir[cirnum] = 1;

            vis[i] = 1; int x = a[i];

            while (!vis[x])

            {

                vis[x] = 1;

                cir[cirnum] ++;

                x = a[x];

            }

        }

    f[0][0][0] = 1;

    for (int w = 1; w <= cirnum; ++w)

    for (int i = sr; i >= 0; --i)

    for (int j = sb; j >= 0; --j)

    for (int k = sg; k >= 0; --k)

    {

        if (i - cir[w] >= 0) f[i][j][k] += f[i-cir[w]][j][k];

        if (f[i][j][k] > yu) f[i][j][k] %= yu;

        if (j - cir[w] >= 0) f[i][j][k] += f[i][j-cir[w]][k];

        if (f[i][j][k] > yu) f[i][j][k] %= yu;

        if (k - cir[w] >= 0) f[i][j][k] += f[i][j][k-cir[w]];

        if (f[i][j][k] > yu) f[i][j][k] %= yu;

    }

    return f[sr][sb][sg];

}

 

int main()

{

    scanf("%d%d%d%d%lld", &sr, &sb, &sg, &m, &yu);

    n = sr+sb+sg;

    int ans = 0;

    for (int i = 1; i <= n; ++i) a[i] = i;

    ans += cal(); if (ans > yu) ans %= yu;

    for (int i = 1; i <= m; ++i)

    {

        for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);

        ans += cal(); if (ans > yu) ans %= yu;

    }

    ans *= mi(m+1, yu-2)%yu;

    printf("%d\n", ans % yu);

}