洛谷 [P2886] 牛继电器Cow Relays

最短路 + 矩阵快速幂

我们可以改进矩阵快速幂,使得它适合本题
用图的邻接矩阵和快速幂实现
注意 dis[i][i] 不能置为 0

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <cstdlib>
using namespace std;
struct edge{
	int u, v, dis;
}e[10000];
int n, m, p, ss, tt;
void work() {
	int sub[10005];
	for(int i = 1; i <= m; i++) {
		cin >> e[i].dis >> e[i].u >> e[i].v;
		sub[++n] = e[i].u;
		sub[++n] = e[i].v;
	}
	sort(sub + 1, sub + n + 1);
	n = unique(sub + 1, sub + n + 1) - sub - 1;
	for(int i = 1; i <= m; i++) {
		e[i].u = lower_bound(sub + 1, sub + n + 1, e[i].u) - sub;
		e[i].v = lower_bound(sub + 1, sub + n + 1, e[i].v) - sub; 
	}
	ss = lower_bound(sub + 1, sub + n + 1, ss) - sub;
	tt = lower_bound(sub + 1, sub + n + 1, tt) - sub;
}
struct Matrix{
	int num[205][205];
	void clear() {
		memset(num, 0x3f, sizeof(num));
	}
	void unit() {
		memset(num, 0, sizeof(num));
		for(int i = 0; i < 205; i++) num[i][i] = 1;
	}
	Matrix operator * (const Matrix & b) {
		Matrix ans;
		ans.clear();
		for(int i = 1; i <= n; i++) {
			for(int j = 1; j <= n; j++) {
				for(int k = 1; k <= n; k++) {
					ans.num[i][j] = min(ans.num[i][j], num[i][k] + b.num[k][j]);
				}
			}
		}
		return ans;
	}
	void print() {
		for(int i = 1; i <= n; i++) {
			for(int j = 1; j <= n; j++) {
				printf("%d ", num[i][j]);
			}
			printf("\n");
		}
	}
	Matrix operator ^ (int k) {
		Matrix ans;
		k--;
		ans = (*this);
		while(k) {
			if(k & 1) ans = ans * (*this);
			(*this) = (*this) * (*this);
			k >>= 1;
		}
		return ans;
	}
};
int main() {
	cin >> p >> m >> ss >> tt;
	work();
	Matrix a;
	a.clear();
	//for(int i = 1; i <= n; i++) a.num[i][i] = 0;
	for(int i = 1; i <= m; i++) {
		a.num[e[i].u][e[i].v] = a.num[e[i].v][e[i].u] = min(e[i].dis, a.num[e[i].u][e[i].v]);
	}
	a = a ^ p;
	//a.print();
	printf("%d\n", a.num[ss][tt]);
	return 0;
}