【SCOI2009】迷路

最新推荐文章于 2025-09-12 17:31:31 发布

转载最新推荐文章于 2025-09-12 17:31:31 发布 · 57 阅读

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原文链接：http://www.cnblogs.com/cj-xxz/p/9879910.html

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#数据结构与算法

题面

题解

如果给我们的是一个邻接矩阵，那么直接给邻接矩阵\(T\)次幂即可。

这里的图有边权，那么我们就将它拆成\(9\)个点即可。

代码

#include<cstdio>
#include<cstring>
#include<algorithm>
#define RG register
#define file(x) freopen(#x".in", "r", stdin);freopen(#x".out", "w", stdout);

inline int read()
{
    int data = 0, w = 1;
    char ch = getchar();
    while(ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
    if(ch == '-') w = -1, ch = getchar();
    while(ch >= '0' && ch <= '9') data = data * 10 + (ch ^ 48), ch = getchar();
    return data * w;
}

inline int read_char()
{
    char ch = getchar();
    while(ch < '0' || ch > '9') ch = getchar();
    return (ch ^ 48);
}

const int maxn(110), Mod(2009);

int n, T;
inline int Num(const int &x, const int &y) { return x + y * n; }

struct Matrix
{
    int a[maxn][maxn], n;
    Matrix(int n = 0) : n(n) { memset(a, 0, sizeof(a)); }

    inline int *operator [] (const int &id) { return a[id]; }
    inline Matrix operator * (const Matrix &b)
    {
        Matrix c(n);
        for(RG int i = 1; i <= n; i++)
            for(RG int j = 1; j <= n; j++)
                for(RG int k = 1; k <= n; k++)
                    c[i][k] = (c[i][k] + a[i][j] * b.a[j][k]) % Mod;
        return c;
    }
};

int main()
{
    n = read(); T = read();
    Matrix S(n * 9), A(n * 9);
    for(RG int i = 1; i <= n; i++)
    {
        for(RG int j = 1; j < 9; j++)
            S[Num(i, j)][Num(i, j - 1)] = 1;
        for(RG int j = 1, x; j <= n; j++)
            if((x = read_char())) S[i][Num(j, x - 1)] = 1;
    }

    for(RG int i = 1; i <= A.n; i++) A[i][i] = 1;
    while(T) { if(T & 1) A = A * S; S = S * S, T >>= 1; }
    printf("%d\n", A[1][n]);
    return 0;
}

转载于:https://www.cnblogs.com/cj-xxz/p/9879910.html