【POJ3613 Cow Relays】（广义矩阵乘法）

题目链接
先离散化，假设有(P)个点
定义矩阵(A_{ij})表示(i)到(j)只经过一条边的最短路，$${(A^{a+b}){ij}=min{1le kle p} { (A^a)_{ik}+(Ab)_{kj} }}$$
(A^{a+b}_{ij})表示(i)到(j)经过((a+b))条边的最短路。
这不就是(ddp)里常用的广义矩阵乘法吗，直接上快速幂即可。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int b[1010], n, m, s, t, cnt, A, B, C;
struct Matrix{
	int a[220][220];
}M;
Matrix operator * (Matrix a, Matrix b){
	Matrix c;
	for(int i = 1; i <= cnt; ++i)
		for(int j = 1; j <= cnt; ++j){
			c.a[i][j] = 1 << 29;
			for(int k = 1; k <= cnt; ++k)
				c.a[i][j] = min(c.a[i][j], a.a[i][k] + b.a[k][j]);
		}
	return c;
}
int main(){
	scanf("%d%d%d%d", &n, &m, &s, &t);
	memset(M.a, 63, sizeof M.a);
	for(int i = 1; i <= m; ++i){
		scanf("%d%d%d", &C, &A, &B);
		if(!b[A]) b[A] = ++cnt;
		if(!b[B]) b[B] = ++cnt;
		M.a[b[A]][b[B]] = M.a[b[B]][b[A]] = C;
	}
	Matrix now = M; --n;
	while(n){
		if(n & 1) now = now * M;
		M = M * M; n >>= 1;
	}
	printf("%d
", now.a[b[s]][b[t]]);
	return 0;
}