HDU - 5411 CRB and Puzzle 矩阵快速幂

HDU - 5411

考虑直接dp会T， 用矩阵优化一下就好了。

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 50 + 2;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 2015;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T &a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T &a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T &a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T &a, S b) {return a > b ? a = b, true : false;}

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int MN, n, m;

struct Matrix {
    int a[N][N];
    Matrix() {
        for(int i = 0; i < MN; i++) {
            for(int j = 0; j < MN; j++) {
                a[i][j] = 0;
            }
        }
    }
    void init() {
        for(int i = 0; i < MN; i++) {
            a[i][i] = 1;
        }
    }
    Matrix operator * (const Matrix &B) {
        Matrix C;
        for(int i = 0; i < MN; i++) {
            for(int j = 0; j < MN; j++) {
                for(int k = 0; k < MN; k++) {
                    C.a[i][j] = (C.a[i][j] + a[i][k] * B.a[k][j]) % mod;
                }
            }
        }
        return C;
    }
    Matrix operator ^ (int b) {
        Matrix A = (*this);
        Matrix C; C.init();
        while(b) {
            if(b & 1) C = C * A;
            A = A * A; b >>= 1;
        }
        return C;
    }
} mat;

int main() {
    int T; scanf("%d", &T);
    while(T--) {
        scanf("%d%d", &n, &m);
        MN = n + 1;
        for(int i = 0; i < MN; i++) {
            for(int j = 0; j < MN; j++) {
                mat.a[i][j] = (i == MN - 1) ? 1 : 0;
            }
        }
        for(int i = 0; i < n; i++) {
            int k, x; scanf("%d", &k);
            while(k--) {
                scanf("%d", &x);
                mat.a[i][x - 1]++;
            }
        }
        Matrix ret = mat ^ m;
        int ans = 0;
        for(int j = 0; j < MN; j++) {
            add(ans, ret.a[MN - 1][j]);
        }
        printf("%d\n", ans);
    }
    return 0;
}

/*
*/

原文地址：https://www.cnblogs.com/CJLHY/p/11156609.html