POJ - 3150 :Cellular Automaton(特殊的矩阵，降维优化)

Posted 2021-01-07 hua-dong

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. The order 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 order m. 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 #1
5 3 1 1
1 2 2 1 2

sample input #2
5 3 1 10
1 2 2 1 2

Sample Output

sample output #1
2 2 2 2 1

sample output #2
2 0 0 2 2

题意：题面很臭很长。大意是，有一个大小为N的环，给出M，K，D，以及N个数。我们进行K次操作，每次操作把距离当前点不超过D的累加到当前点，结果模M。

思路：因为要进行K次，每次的原则是一样的，我们可以想到用矩阵来优化，如果i能到达j，把么base[i][j]=1；则结果ans=A*(base^K)。

但是需要优化，时间复杂度为O(N^3*lgK)。我们发现矩阵是下一行由上一行右移一位而来，那么我们保存一维即可代表这个矩阵。同样的，我们只需要得到第一行的矩阵结果，就能得到整个矩阵的结果。

#include<bits/stdc++.h>
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=510;
int N,Mod,D,K;
struct mat
{
    int M[maxn];
    mat(){ rep(i,1,N) M[i]=0; }
    mat friend operator*(mat a,mat b){
        mat res;
        rep(k,1,N)
         rep(j,1,N){
            res.M[j]=(res.M[j]+(ll)a.M[k]*b.M[j-k+1>0?j-k+1:j-k+1+N]%Mod)%Mod;
         }
        return res;
    }
    mat friend operator ^(mat a,int x)
    {
       mat res;rep(i,1,N) res.M[1]=1;
       while(x){
          if(x&1) res=res*a; a=a*a; x/=2;
       } return res;
    }

};
int main()
{
    scanf("%d%d%d%d",&N,&Mod,&D,&K);
    mat a,base;
    rep(i,1,N) scanf("%d",&a.M[i]);
    rep(i,1,N)
      if(i-1<=D||N-i+1<=D||N-1+i<=D) base.M[i]=1;
    a=a*(base^K);
    rep(i,1,N-1) printf("%d ",a.M[i]);
    printf("%d
",a.M[N]);
    return 0;
}