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  • bzoj 1875 矩阵快速幂

    思路:不能走走过来的路,变点交换跑矩阵快速幂。

    #include<bits/stdc++.h>
    #define LL long long
    #define fi first
    #define se second
    #define mk make_pair
    #define PII pair<int, int>
    #define y1 skldjfskldjg
    #define y2 skldfjsklejg
    using namespace std;
    
    const int N = 1e5 + 7;
    const int M = 5e5 + 7;
    const int inf = 0x3f3f3f3f;
    const LL INF = 0x3f3f3f3f3f3f3f3f;
    const int mod = 45989;
    
    int n, m, t, S, T, U[N], V[N], b[N];
    
    struct Matrix {
        int a[125][125], n;
    
        Matrix(int _n) {
            n = _n;
            memset(a, 0, sizeof(a));
        }
    
        void init() {
            for(int i = 0; i < n; i++)
                a[i][i] = 1;
        }
    
        Matrix operator * (const Matrix &B) const {
            Matrix C(n);
            for(int i = 0; i < n; i++)
                for(int j = 0; j < n; j++)
                    for(int k = 0; k < n; 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 ans(n);
            ans.init();
            while(b) {
                if(b & 1) ans = ans * A;
                A = A * A; b >>= 1;
            }
            return ans;
        }
    };
    int main() {
        scanf("%d%d%d%d%d", &n, &m, &t, &S, &T);
        for(int i = 1; i <= m; i++) {
            scanf("%d%d", &U[i << 1], &V[i << 1]);
            U[i << 1 | 1] = V[i << 1];
            V[i << 1 | 1] = U[i << 1];
        }
        m = m << 1 | 1;
        Matrix A(m + 1);
        for(int i = 2; i <= m; i++) {
            if(U[i] == S) A.a[0][i] = 1;
            if(V[i] == T) A.a[i][1] = 1;
            for(int j = 2; j <= m; j++)
                if(V[i] == U[j] && (i ^ j) != 1)
                    A.a[i][j] = 1;
        }
        A = A ^ (t + 1);
        printf("%d
    ", A.a[0][1]);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/CJLHY/p/9599720.html
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