51Nod-1032-骨牌覆盖 V2

ACM模版

描述

题解

数据弱化的一个题，原题是 51Nod1033骨牌覆盖V2，插头 DP。

代码

#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>using namespace std;typedef long long ll;const int MOD = 1e9 + 7;const int MAXN = 1 << 5;ll dp[MAXN][MAXN];ll ret[MAXN][MAXN];ll tmp[MAXN][MAXN];int m, n = 3;void dfs(int col, int pre, int now){    if (col > n)    {        return ;    }    if (col == n)    {        dp[pre][now]++;        return ;    }    dfs(col + 1, pre << 1, (now << 1) | 1);    dfs(col + 1, (pre << 1) | 1, now << 1);    dfs(col + 2, pre << 2 , now << 2);}void mul(ll ret[][MAXN], ll a[][MAXN], ll b[][MAXN]){    int t = 1 << n;    for (int i = 0; i < t; i++)    {        for (int j = 0; j < t; j++)        {            ll tmp = 0;            for (int k = 0; k < t; k++)            {                tmp += a[i][k] * b[k][j];                tmp %= MOD;            }            ret[i][j] = tmp;        }    }}int main(){    scanf("%d", &m);    dfs(0, 0, 0);    int t = 1 << n;    for (int i = 0; i < t; i++)    {        ret[i][i] = 1;    }    m++;    while (m)    {        for (int i = 0; i < t; i++)        {            for (int j = 0; j < t; j++)            {                tmp[i][j] = ret[i][j];            }        }        if (m & 1)        {            mul(ret, tmp, dp);        }        m = m >> 1;        mul(tmp, dp, dp);        for (int i = 0; i < t; i++)        {            for (int j = 0; j < t; j++)            {                dp[i][j] = tmp[i][j];            }        }    }    cout << ret[0][t - 1] << endl;    return 0;}