牛客小白月赛7 方格填色

最新推荐文章于 2024-09-10 19:26:26 发布

转载最新推荐文章于 2024-09-10 19:26:26 发布 · 234 阅读

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原文链接：http://www.cnblogs.com/widsom/p/9702981.html

本文介绍了一种使用矩阵快速幂优化动态规划的方法，解决了一个方格填色问题。通过构建特定矩阵并利用快速幂运算，可以高效地计算出n行m列的方格中所有合法填色方案的数量。

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方格填色

思路：

用矩阵快速幂优化dp

代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL unsigned long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pii pair<int, int>
#define piii pair<pii, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int MOD = 1e9 + 7;
const int M = 1<<5;
int m;
LL n;
struct Matrix {
    LL a[M][M];
    void init() {
        for (int i = 0; i < (1<<m); i++) {
            for (int j = 0; j < (1<<m); j++) 
                a[i][j] = 0;
        }
    }
    void _init() {
        init();
        for (int i = 0; i < (1<<m); i++) {
            a[i][i] = 1; 
        }
    }
}A;
Matrix mul(Matrix A, Matrix B) {
    Matrix ans;
    ans.init();
    for (int i = 0; i < (1<<m); i++) {
        for (int j = 0; j < (1<<m); j++) {
            if(A.a[i][j]) {
                for (int k = 0; k < (1<<m); k++) 
                    ans.a[i][k] = (ans.a[i][k] + A.a[i][j] * B.a[j][k]) % MOD;
            }
        }
    }
    return ans;
} 
Matrix q_pow(Matrix A, LL k) {
    Matrix ans;
    ans._init();
    while(k) {
        if(k&1) ans = mul(ans, A);
        A = mul(A, A);
        k >>= 1; 
    }
    return ans;
} 
int main() {
    scanf("%d %lld", &m, &n);
    for (int i = 0; i < (1<<m); i++) {
        for (int j = 0; j < (1<<m); j++) {
            bool f = true;
            for (int k = 0; k < m; k++) {
                if(!(i&(1<<k)) && !(j&(1<<k))) {
                    f = false;
                    break;
                }
            }
            if(i == (1<<m)-1 && j == i) f = false;
            A.a[i][j] = f;
        }
    }
    A = q_pow(A, n-1);
    LL ans = 0;
    for (int i = 0; i < (1<<m); i++) {
        for (int j = 0; j < (1<<m); j++) {
            ans = (ans + A.a[i][j]) % MOD;
        }
    }
    printf("%lld\n", ans);
    return 0;
}