[Sdoi 2009] HH for a walk

[Question link]

Https://www.lydsy.com/JudgeOnline/problem.php? Id = 1875

[Algorithm]

Use F [I] [J] to show the number of solutions on the J edge while taking step I.

Matrix acceleration.

Time Complexity: O (N ^ 3 logn)

[Code]

#include<bits/stdc++.h>using namespace std;#define MAXN 125const int P = 45989;struct edge{        int to , nxt;} e[MAXN << 1];int n , m , t , A , B , tot;int head[MAXN];struct matrix_t{        int mat[MAXN][MAXN];    } a , b;template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }template <typename T> inline void read(T &x){    T f = 1; x = 0;    char c = getchar();    for (; !isdigit(c); c = getchar()) if (c == ‘-‘) f = -f;    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - ‘0‘;    x *= f;}inline void addedge(int u,int v){        tot++;        e[tot] = (edge){v , head[u]};        head[u] = tot;}inline void multipy(matrix_t &a , matrix_t b){        static matrix_t ret;        for (int i = 1; i <= tot; i++)        {                for (int j = 1; j <= tot; j++)                {                        ret.mat[i][j] = 0;                }        }        for (int i = 1; i <= tot; i++)        {                for (int j = 1; j <= tot; j++)                {                        for (int k = 1; k <= tot; k++)                        {                                ret.mat[i][j] = (ret.mat[i][j] + 1LL * a.mat[i][k] * b.mat[k][j]) % P;                         }                }        }        a = ret;}inline void exp_mod(matrix_t &a , int t){        static matrix_t tmp;        for (int i = 1; i <= tot; i++)        {                for (int j = 1; j <= tot; j++)                {                        tmp.mat[i][j] = (i == j);                }        }        while (t > 0)        {                if (t & 1) multipy(tmp , a);                multipy(a , a);                t >>= 1;        }                a = tmp;}int main(){                read(n); read(m); read(t); read(A); read(B);        ++A; ++B;        tot = 1;        for (int i = 1; i <= m; i++)        {                int u , v;                read(u); read(v);                ++u; ++v;                addedge(u , v);                addedge(v , u);        }        for (int i = 2; i <= tot; i++)        {                for (int j = 2; j <= tot; j++)                {                        if (i != (j ^ 1) && e[j ^ 1].to == e[i].to)                                ++b.mat[i][j];                                }        }        for (int i = head[A]; i; i = e[i].nxt) ++a.mat[1][i];        exp_mod(b , t - 1);        multipy(a , b);        int ans = 0;        for (int i = head[B]; i; i = e[i].nxt) ans = (ans + 1LL * a.mat[1][i ^ 1]) % P;        printf("%d\n" , ans);                return 0;    }

 

[Sdoi 2009] HH for a walk