1120考试T2

1120考试T2

​ 题目大意:

​ 给你一个n个点，m条边的图，每条边有一个权值在某几个时刻有一些点不能走，每次必须走一条边，求恰好k时刻到达终点的路径中最大边权最小.

​ 其中不能走的点是这样的: 有(num)个机器人, 每个机器人会周期性的按顺序走一些点, 编号为(t_1, t_2, ...)

​ (n <= 50, m <= 1600, k <= 1e8, num <= 10, t_i <= 4)

​ 与这道题很像, 只不过这道题求方案数

​ 矩阵快速幂.

​ 我们发现(t)最大只有4, 得出这些机器人最多只会有12种所在点的状态, 也就是12个一循环.

​ 那么我们可以处理出这12个矩阵, 也就是在这些状态下那些点不能走, 然后把这12个矩阵相乘, 就得到了一次循环后, 各点之间路径的最大边权最小值.

​ 然后我们把这个乘后的矩阵作为转移矩阵, 用矩阵快速幂, 那么指数就应该是(k / 12).

​ 对于剩下的(k \% 12)次暴力乘就好了.

​ 复杂度(O(n^3log k))

#include <bits/stdc++.h>

using namespace std;

inline long long read() {
	long long s = 0, f = 1; char ch;
	while(!isdigit(ch = getchar())) (ch == '-') && (f = -f);
	for(s = ch ^ 48;isdigit(ch = getchar()); s = (s << 1) + (s << 3) + (ch ^ 48));
	return s * f;
}

const int N = 55, inf = 1e9;
int n, m, k, s, t, num;
int a[N], now[N], dis[N][N], num_a[N];
struct mat { 
	int v[N][N]; 
	void init() { for(int i = 1;i <= n; i++) for(int j = 1;j <= n; j++) v[i][j] = inf; }
	void with_dis() { for(int i = 1;i <= n; i++) for(int j = 1;j <= n; j++) v[i][j] = dis[i][j]; }
} ans, turn, q[13];

mat operator * (const mat &a, const mat &b) {
	mat c; c.init();
	for(int i = 1;i <= n; i++)
		for(int j = 1;j <= n; j++)
			for(int k = 1;k <= n; k++) 
				if(a.v[i][k]) c.v[i][j] = min(c.v[i][j], max(a.v[i][k], b.v[k][j]));
	return c;
}

mat ksm(mat x, int y) {
	mat res; res.init();
	for(int i = 1;i <= n; i++) res.v[i][i] = 1;
	while(y) { if(y & 1) res = res * x; x = x * x; y >>= 1; }
	return res;
}

int main() {

	n = read(); m = read(); k = read();
	if(m == 0 && k == 0) { printf("0"); return 0; }
	for(int i = 1;i <= n; i++)
		for(int j = 1;j <= n; j++) dis[i][j] = inf;
	for(int i = 1, x, y, z;i <= m; i++) {
		x = read(); y = read(); z = read();
		if(dis[x][y]) dis[x][y] = dis[y][x] = min(dis[x][y], z);
		else dis[x][y] = dis[y][x] = z;
	}
	for(int i = 1;i <= 12; i++) q[i].with_dis();
	num = read(); s = read(); t = read(); ans.init();
	for(int i = 1, x;i <= num; i++) {
		x = read();
		for(int j = 1;j <= x; j++) a[j] = read();
		for(int j = 1;j <= 12; j++) {
			int p = a[j % x == 0 ? x : j % x];
			for(int l = 1;l <= n; l++) q[j].v[l][p] = inf;
		}
	}
	turn = q[1];
	for(int i = 2;i <= 12; i++) turn = turn * q[i];
	ans = ksm(turn, k / 12);
	for(int i = 1;i <= k % 12; i++) ans = ans * q[i];
	if(ans.v[s][t] == 1e9) printf("impossible");
	else printf("%d", ans.v[s][t]);

	return 0;
}