VIJOS 1477 跳动的水珠

最新推荐文章于 2019-11-04 19:20:17 发布

原创最新推荐文章于 2019-11-04 19:20:17 发布 · 1k 阅读

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#置换 #快速幂

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本文介绍了一种使用矩阵快速幂和倍增置换方法解决特定问题的高效算法。该算法通过优化循环节来加速计算过程，避免了传统矩阵方法的资源浪费。文章详细解释了如何构建和运用转移矩阵来迭代计算，并提供了完整的代码实现。

提示：
1. 本题的暴力可以先写好备着，总是有用的。
2. 看到T的范围，我打赌跟快速幂有关……

详细题解和分析在代码后：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include <vector>
#include <deque>
#include <set>
#include <map>
#include <complex>
#include <climits>
#include <cassert>

using namespace std;
const int maxn = 120;
const int maxm = 102;
int modu;

typedef long long ll;
struct Matrix
{
    int n , m;
    int a[maxm][maxm];

    Matrix(int n=0 , int m=0){ this->n = n; this->m = m; memset(a , 0 , sizeof a); }

    int* operator [] ( int b ) { return a[b]; }
    Matrix operator *(Matrix b)
    {
        Matrix res(n , b.m);
        assert(m == b.n);

        for(int i=1;i<=n;i++) for(int j=1;j<=b.m;j++) for(int k=1;k<=m;k++) (res[i][j] += a[i][k]*b[k][j]) %= modu;
        return res;
    }
};

Matrix powMatrix(Matrix a , int n)
{
    Matrix res;
    res.n = res.m = a.n;

    for(int i=1;i<=res.n;i++) res[i][i] = 1;

    while(n)
    {
        if(n&1) res = res*a;
        a = a*a;
        n >>= 1;
    }

    return res;
}

int n , m , x , p , t;
string d[maxn][maxn];

Matrix timeFly(Matrix a , int t)
{
    int b[maxn][maxn];
    for(int i=0;i<t;i++)
    {
        memset(b , 0 , sizeof b);
        for(int j=1;j<=n;j++) for(int k=1;k<=m;k++)
        {
            char c = d[j][k][i%d[j][k].size()];
            if(c == 'C') (b[j][k] += (x+a[j][k])%p) %= p;
            if(c == 'U') (b[j-1][k] += a[j][k]) %= p;
            if(c == 'D') (b[j+1][k] += a[j][k]) %= p;
            if(c == 'L') (b[j][k-1] += a[j][k]) %= p;
            if(c == 'R') (b[j][k+1] += a[j][k]) %= p;
        }

        for(int j=1;j<=n;j++) for(int k=1;k<=m;k++) a[j][k] = b[j][k];
    }
    return a;
}

void print(Matrix a)
{
    for(int i=1;i<=n;i++)
    {
        for(int j=1,k;j<=m;j=k)
        {
            for(k=j;k<=m && a[i][k] == a[i][j];k++);
            cout<<(k-j)<<" "<<a[i][j]<<(k>m?"\n":" ");
        }
    }
}

int id(int x , int y) { return (x-1)*m + y; }

__inline Matrix modifyTrans(Matrix a , Matrix b)
{
    Matrix res;
    for(int i=1;i<=n;i++) for(int j=1;j<=m;j++)if(a[i][j])
    {
        int x = (a[i][j]+m-1)/m , y = (a[i][j]%m==0)?m:a[i][j]%m;
        res[i][j] = b[x][y];
    }

    return res;
}

__inline Matrix modifyNow(Matrix a , Matrix b , Matrix c)
{
    Matrix res = c;
    for(int i=1;i<=n;i++) for(int j=1;j<=m;j++)if(b[i][j])
    {
        int x = (b[i][j]+m-1)/m , y = (b[i][j]%m==0)?m:b[i][j]%m;
        (res[x][y] += a[i][j]) %= modu;
    }

    return res;
}


int main()
{


    cin>>n>>m>>x>>p>>t;
    modu = p;

    for(int i=1;i<=n;i++) for(int j=1,k;j<=m;j++) cin>>k>>d[i][j];
    //puts("OK");
    /*if(t <= 1e4)
    {
        print(timeFly(Matrix() , t));
        return 0;
    }
    else */ 
    {
        //puts("OK");
        Matrix now = timeFly(Matrix() , 63) , res;

        Matrix trans(n , m);
        for(int i=1;i<=n;i++) for(int j=1;j<=m;j++)
        {
            int x = i , y = j; bool ok = true;
            for(int k=0,nx,ny;k<63;k++) 
            {
                char c = d[x][y][k%d[x][y].size()];
                if(c == 'C') nx = x , ny = y;
                if(c == 'U') nx = x-1 , ny = y;
                if(c == 'D') nx = x+1 , ny = y;
                if(c == 'L') nx = x , ny = y-1;
                if(c == 'R') nx = x , ny = y+1;
                x = nx;
                y = ny;
                if(!x || x>n || !y || y>m) { ok = false; break; }
            }
            if(ok) trans[i][j] = id(x , y);
        }

        int req = t/63;
        while(req)
        {
            if(req&1) res = modifyNow(res , trans , now);
            now = modifyNow(now , trans , now);
            trans = modifyTrans(trans , trans);

            req >>= 1;
        }

        print(timeFly(res , t%63));
    }


    return 0;
}