[POJ 3150] Cellular Automaton (矩阵快速幂 + 矩阵乘法优化）

Cellular Automaton

Time Limit: 12000MS Memory Limit: 65536KTotal Submissions: 3048 Accepted: 1227Case Time Limit: 2000MS

Description

A cellular automaton is a collection of cells on a grid of specified shape that evolves through a number of discrete time steps according to a set of rules that describe the new state of a cell based on the states of neighboring cells. Theorder of the cellular automaton is the number of cells it contains. Cells of the automaton of order n are numbered from 1 to n.

The order of the cell is the number of different values it may contain. Usually, values of a cell of order m are considered to be integer numbers from 0 to m − 1.

One of the most fundamental properties of a cellular automaton is the type of grid on which it is computed. In this problem we examine the special kind of cellular automaton — circular cellular automaton of order n with cells of orderm. We will denote such kind of cellular automaton as n,m-automaton.

A distance between cells i and j in n,m-automaton is defined as min(|i − j|, n − |i − j|). A d-environment of a cell is the set of cells at a distance not greater than d.

On each d-step values of all cells are simultaneously replaced by new values. The new value of cell i after d-step is computed as a sum of values of cells belonging to the d-enviroment of the cell i modulo m.

The following picture shows 1-step of the 5,3-automaton.

The problem is to calculate the state of the n,m-automaton after k d-steps.

Input

The first line of the input file contains four integer numbers n, m, d, and k (1 ≤ n ≤ 500, 1 ≤ m ≤ 1 000 000, 0 ≤ d < ⁿ⁄₂ , 1 ≤ k ≤ 10 000 000). The second line contains n integer numbers from 0 to m − 1 — initial values of the automaton’s cells.

Output

Output the values of the n,m-automaton’s cells after k d-steps.

Sample Input

sample input #15 3 1 11 2 2 1 2sample input #25 3 1 101 2 2 1 2

Sample Output

sample output #12 2 2 2 1sample output #22 0 0 2 2

Source

Northeastern Europe 2006

题目大意：

有一个环，长度为n。其中每一个结点上面都有数字。可以对环上的每一个结点进行更新，使得该结点的值变为环上距离该结点的距离小于等于d的所有节点数之和，最后再对m取模。每次操作更新一遍环上所有点，问经过k次操作之后，环上各点数字为多少。

解题思路：

　　首先可以想到用矩阵来代替操作。（对于题目第一组样例）

首先矩阵a = 1 2 2 1 2 
b = 
1 1 0 0 1
1 1 1 0 0
0 1 1 1 0
0 0 1 1 1
1 0 0 1 1

然后可以通过矩阵快速幂来解决。这样复杂度为（logk * n^3），会T。

其实这个矩阵是有规律的。

b^1 =
[1, 1, 0, 0, 1]
[1, 1, 1, 0, 0]
[0, 1, 1, 1, 0]
[0, 0, 1, 1, 1]
[1, 0, 0, 1, 1]
b^2 =
[3, 2, 1, 1, 2]
[2, 3, 2, 1, 1]
[1, 2, 3, 2, 1]
[1, 1, 2, 3, 2]
[2, 1, 1, 2, 3]
b^3 =
[7, 6, 4, 4, 6]
[6, 7, 6, 4, 4]
[4, 6, 7, 6, 4]
[4, 4, 6, 7, 6]
[6, 4, 4, 6, 7]
b^4 =
[19, 17, 14, 14, 17]
[17, 19, 17, 14, 14]
[14, 17, 19, 17, 14]
[14, 14, 17, 19, 17]
[17, 14, 14, 17, 19]

观察可以发现，只要把矩阵第i行的第一个数字移到最后，就变成了矩阵的第 i + 1 行。所以我们只需要知道矩阵的某一行，就可以推得其他行。所以N^3的复杂度就将为N^2。

代码：

/*ID: [e-mail]:LANG: C++*/#include<map>#include<set>#include<queue>#include<stack>#include<cmath>#include<cstdio>#include<vector>#include<string>#include<fstream>#include<cstring>#include<ctype.h>#include<iostream>#include<algorithm>#define INF (1<<30)#define PI acos(-1.0)#define mem(a, b) memset(a, b, sizeof(a))#define For(i, n) for (int i = 0; i < n; i++)typedef long long ll;using namespace std;const int maxn = 505;const int maxm = 505;int mod, n, k, d;void Matrix_pow(int a[], int b[]) {    ll c[505] = {0};    for (int i = 0; i < n; i++) {        for (int j = 0; j < n; j++) c[i] += ((ll)a[j] * b[(i + j) % n]) % mod;        c[i] %= mod;    }    for (int i = 0; i < n; i++) b[i] = c[i];}int a[505], b[505];int main () {    while(scanf("%d%d%d%d", &n, &mod, &d, &k) != EOF) {        for (int i = 0; i < n; i++) scanf("%d", b + i);        memset(a, 0, sizeof(a));        a[0] = 1;        for (int i = 1; i <= d; i++) a[i] = a[n - i] = 1;        while(k) {            if (k & 1) Matrix_pow(a, b);            k >>= 1;            Matrix_pow(a, a);        }        for (int i = 0; i < n; i++) printf("%d%c", b[i], " \n"[i == n - 1]);    }    return 0;}

参考：http://www.cppblog.com/varg-vikernes/archive/2011/02/08/139804.html