bzoj1297 [SCOI2009]迷路 矩阵快速幂

题目传送门

https://lydsy.com/JudgeOnline/problem.php?id=1297

题解

如果每一条边没有边权，那么就是一般的矩阵快速幂。

因为边权 \(\in [1, 9]\)，所以我们可以把每一个点拆成 \(9\) 个点，对于一条边权为 \(w\) 的边 \(x \to y\)，可以建立一条边 \(x_{w-1}\to y_0\)。

然后 \(x_i\to x_{i+1}\) 连边。

然后矩阵快速幂就可以了。

#include<bits/stdc++.h>

#define fec(i, x, y) (int i = head[x], y = g[i].to; i; i = g[i].ne, y = g[i].to)
#define dbg(...) fprintf(stderr, __VA_ARGS__)
#define File(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
#define fi first
#define se second
#define pb push_back

template<typename A, typename B> inline char smax(A &a, const B &b) {return a < b ? a = b, 1 : 0;}
template<typename A, typename B> inline char smin(A &a, const B &b) {return b < a ? a = b, 1 : 0;}

typedef long long ll; typedef unsigned long long ull; typedef std::pair<int, int> pii;

template<typename I> inline void read(I &x) {
    int f = 0, c;
    while (!isdigit(c = getchar())) c == '-' ? f = 1 : 0;
    x = c & 15;
    while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + (c & 15);
    f ? x = -x : 0;
}

const int N = 90 + 3;
const int P = 2009;

int n, m, T;
char s[N];

inline int smod(int x) { return x >= P ? x - P : x; }
inline void sadd(int &x, const int &y) { x += y; x >= P ? x -= P : x; }
inline int fpow(int x, int y) {
    int ans = 1;
    for (; y; y >>= 1, x = x * x % P) if (y & 1) ans = ans * x % P;
    return ans;
}

struct Matrix {
    int a[N][N];
    
    inline Matrix() { memset(a, 0, sizeof(a)); }
    inline Matrix(const int &x) {
        memset(a, 0, sizeof(a));
        for (int i = 1; i <= m; ++i) a[i][i] = x;
    }
    
    inline Matrix operator * (const Matrix &b) {
        Matrix c;
        for (int k = 1; k <= m; ++k)
            for (int i = 1; i <= m; ++i)
                for (int j = 1; j <= m; ++j)
                    sadd(c.a[i][j], a[i][k] * b.a[k][j] % P);
        return c;
    }
} A;

inline Matrix fpow(Matrix x, int y) {
    Matrix ans(1);
    for (; y; y >>= 1, x = x * x) if (y & 1) ans = ans * x;
    return ans;
}

inline void work() {
    m = 9 * n;
    A = fpow(A, T);
    printf("%d\n", A.a[1][n]);
}

inline void init() {
    read(n), read(T);
    for (int i = 1; i <= n; ++i) {
        scanf("%s", s + 1);
        for (int j = 1; j <= n; ++j)
            if (s[j] != '0') A.a[i + (s[j] - '1') * n][j] = 1;
    }
    for (int i = 1; i <= n; ++i)
        for (int j = 0; j < 9 - 1; ++j)
            A.a[i + j * n][i + (j + 1) * n] = 1;
}

int main() {
#ifdef hzhkk
    freopen("hkk.in", "r", stdin);
#endif
    init();
    work();
    fclose(stdin), fclose(stdout);
    return 0;
}

标签：const,迷路,return,int,bzoj1297,SCOI2009,ans,inline,define
来源： https://www.cnblogs.com/hankeke/p/bzoj1297.html