沼泽鳄鱼

ssl 2511

题目大意

给你一个无向图，有一些鳄鱼有周期性地在这个图中走（鳄鱼不用沿着边走，周期为2或3或4），问你从初始点走到最终点走k个单位时间，不在点和边上停下，不在同一时间和鳄鱼在同一点，有多少种走法

输入样例

6 8 1 5 3
0 2
2 1
1 0
0 5
5 1
1 4
4 3
3 5
1
3 0 5 1

输出样例

2

样例解释

时刻 -----------0 1 2 3
食人鱼位置 --0 5 1 0
路线一 --------1 2 0 5
路线二 --------1 4 3 5

数据范围

$1⩽N⩽50$
$1⩽K⩽2,000,000,000$
$1⩽NFish⩽20$

解题思路

可以拿2,3,4的最小公倍数12当成一个单位时间
然后预处理出12个时间的矩阵
然后快矩阵速幂
最后把余数处理一下即可

代码

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define wyc 10000
using namespace std;
int n, m, t, x, y, st, ed, fn, s[100], v[100][100];
struct matrix
{int n, m, a[60][60];matrix operator *(matrix const &b) const{matrix c;c.n = n;c.m = b.m;for (int i = 0; i < c.n; ++i)for (int j = 0; j < c.m; ++j)c.a[i][j] = 0;for (int i = 0; i < c.n; ++i)for (int k = 0; k < m; ++k)for (int j = 0; j < c.m; ++j)c.a[i][j] = (c.a[i][j] + a[i][k] * b.a[k][j] % wyc) % wyc;return c;}
}A, B, C;
matrix solve(matrix b, int x)//预处理
{matrix a;for (int i = 1; i <= x; ++i){matrix c = b;for (int j = 1; j <= fn; ++j)for (int k = 0; k < n; ++k)c.a[k][v[j][i % s[j]]] = 0;//有鳄鱼的地方直接清空if (i > 1) a = a * c;else a = c;}return a;
}
void Counting(int x)
{while(x){if (x&1) A = A * B;B = B * B;x>>=1;}return;
}
int main()
{scanf("%d%d%d%d%d", &n, &m, &st, &ed, &t);C.n = C.m = A.m = n;A.n = 1;A.a[0][st] = 1;for (int i = 1; i <= m; ++i){scanf("%d%d", &x, &y);C.a[x][y] = 1;C.a[y][x] = 1;}scanf("%d", &fn);for (int i = 1; i <= fn; ++i){scanf("%d", &s[i]);for (int j = 0; j < s[i]; ++j)scanf("%d", &v[i][j]);}B = solve(C, 12);Counting(t / 12);if (t % 12) A = A * solve(C, t % 12);//解决余数printf("%d", A.a[0][ed]);return 0;
}