牛客小白月赛7 方格填色

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;
}

 

转载于:https://www.cnblogs.com/widsom/p/9702981.html

### 关于牛客小白109的信息 目前并未找到关于牛客小白109的具体比信息或题解内容[^5]。然而,可以推测该事可能属于牛客网举办的系列算法竞之一,通常这类比会涉及数据结构、动态规划、图论等经典算法问题。 如果要准备类似的事,可以通过分析其他场次的比题目来提升自己的能力。例如,在牛客小白13中,有一道与二叉树相关的题目,其核心在于处理树的操作以及统计最终的结果[^3]。通过研究此类问题的解决方法,能够帮助理解如何高效地设计算法并优化时间复杂度。 以下是基于已有经验的一个通用解决方案框架用于应对类似场景下的批量更新操作: ```python class TreeNode: def __init__(self, id): self.id = id self.weight = 0 self.children = [] def build_tree(n): nodes = [TreeNode(i) for i in range(1, n + 1)] for node in nodes: if 2 * node.id <= n: node.children.append(nodes[2 * node.id - 1]) if 2 * node.id + 1 <= n: node.children.append(nodes[2 * node.id]) return nodes[0] def apply_operations(root, operations, m): from collections import defaultdict counts = defaultdict(int) def update_subtree(node, delta): stack = [node] while stack: current = stack.pop() current.weight += delta counts[current.weight] += 1 for child in current.children: stack.append(child) def exclude_subtree(node, total_nodes, delta): nonlocal root stack = [(root, False)] # (current_node, visited) subtree_size = set() while stack: current, visited = stack.pop() if not visited and current != node: stack.append((current, True)) for child in current.children: stack.append((child, False)) elif visited or current == node: if current != node: subtree_size.add(current.id) all_ids = {i for i in range(1, total_nodes + 1)} outside_ids = all_ids.difference(subtree_size.union({node.id})) for idx in outside_ids: nodes[idx].weight += delta counts[nodes[idx].weight] += 1 global nodes nodes = {} queue = [root] while queue: curr = queue.pop(0) nodes[curr.id] = curr for c in curr.children: queue.append(c) for operation in operations: op_type, x = operation.split(' ') x = int(x) target_node = nodes.get(x, None) if not target_node: continue if op_type == '1': update_subtree(target_node, 1) elif op_type == '2' and target_node is not None: exclude_subtree(target_node, n, 1) elif op_type == '3': path_to_root = [] temp = target_node while temp: path_to_root.append(temp) if temp.id % 2 == 0: parent_id = temp.id // 2 else: parent_id = (temp.id - 1) // 2 if parent_id >= 1: temp = nodes[parent_id] else: break for p in path_to_root: p.weight += 1 counts[p.weight] += 1 elif op_type == '4': pass # Implement similarly to other cases. result = [counts[i] for i in range(m + 1)] return result ``` 上述代码片段展示了针对特定类型的树形结构及其操作的一种实现方式。尽管它并非直接对应小白109中的具体题目,但它提供了一个可借鉴的设计思路。 ####
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