bzoj1706 [usaco2007 Nov]relays 奶牛接力跑 矩阵快速幂

题目传送门

https://lydsy.com/JudgeOnline/problem.php?id=1706

题解

换个方法定义矩阵乘法：先加再取 \(\min\)。

对于一个 \(n\times m\) 的矩阵 \(A\)，和一个 \(m\times l\) 的矩阵 \(B\) 它们的乘积 \(C\) 是一个 \(n \times l\) 的矩阵。

\[C_{i, j} = \min_{k=1}^m A_{i, k}+B_{k,j} \]

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关于这个东西的结合律的证明和一般的矩阵乘法类似，直接带入就可以了。大家可以看一下我的另一篇博客。动态 DP 学习笔记 里面有提到。

然后显然就是先建出来邻接矩阵，然后求它的 \(n\) 次方，这个就是个矩阵快速幂了。

#include<bits/stdc++.h>

#define fec(i, x, y) (int i = head[x], y = g[i].to; i; i = g[i].ne, y = g[i].to)
#define dbg(...) fprintf(stderr, __VA_ARGS__)
#define File(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
#define fi first
#define se second
#define pb push_back

template<typename A, typename B> inline char smax(A &a, const B &b) {return a < b ? a = b, 1 : 0;}
template<typename A, typename B> inline char smin(A &a, const B &b) {return b < a ? a = b, 1 : 0;}

typedef long long ll; typedef unsigned long long ull; typedef std::pair<int, int> pii;

template<typename I> inline void read(I &x) {
	int f = 0, c;
	while (!isdigit(c = getchar())) c == '-' ? f = 1 : 0;
	x = c & 15;
	while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + (c & 15);
	f ? x = -x : 0;
}

const int N = 200 + 7;

int T, n, m, st, ed;
int b[N];

struct Edges { int u, v, w; } a[N];

struct Matrix {
	int a[N][N];
	
	inline Matrix() { memset(a, 0x3f, sizeof(a)); }
	inline Matrix(const int &x) {
		memset(a, 0x3f, sizeof(a));
		for (int i = 1; i <= n; ++i) a[i][i] = x;
	}
	
	inline Matrix operator * (const Matrix &b) {
		Matrix c;
		for (int k = 1; k <= n; ++k)
			for (int i = 1; i <= n; ++i)
				for (int j = 1; j <= n; ++j)
					smin(c.a[i][j], a[i][k] + b.a[k][j]);
		return c;
	}
} A;

inline Matrix fpow(Matrix x, int y) {
	Matrix ans(0);
	for (; y; y >>= 1, x = x * x) if (y & 1) ans = ans * x;
	return ans;
}

inline void work() {
	std::sort(b + 1, b + (m << 1) + 1);
	n = std::unique(b + 1, b + (m << 1) + 1) - b - 1;
	for (int i = 1; i <= m; ++i) {
		int x = a[i].u, y = a[i].v, z = a[i].w;
		x = std::lower_bound(b + 1, b + n + 1, x) - b;
		y = std::lower_bound(b + 1, b + n + 1, y) - b;
		A.a[x][y] = A.a[y][x] = z;
	}
	
	printf("%d\n", fpow(A, T).a[std::lower_bound(b + 1, b + n + 1, st) - b][std::lower_bound(b + 1, b + n + 1, ed) - b]);
}

inline void init() {
	read(T), read(m), read(st), read(ed);
	for (int i = 1; i <= m; ++i)
		read(a[i].w), read(a[i].u), read(a[i].v),
		b[(i << 1) - 1] = a[i].u, b[i << 1] = a[i].v;
}

int main() {
#ifdef hzhkk
	freopen("hkk.in", "r", stdin);
#endif
	init();
	work();
	fclose(stdin), fclose(stdout);
	return 0;
}