[POJ3613]Cow Relays && Floyd倍增

怎么说呢 如果有看论文（矩阵乘法-俞华程） 这道题还是相当异常非常的简单 （除了本人自己玩自己忘记离散化 直接函数爆栈调N久甚至求助吴老师之外） 

唯一需要注意的就是单位矩阵的建立是对角线为0 其他全部为正无穷 

Floyd倍增的原因论文上有说 其实我也没怎么看

废话就说到这里 上代码

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
using namespace std;
#define MAXN 100
#define MAXM 100
#define INF 0x3f3f3f3f
typedef int LL;
struct Matrix{  
    int n, m;  
    LL A[MAXN+10][MAXN+10];  
}E, Z, A;
int Map[MAXN+10][MAXN+10];
int n, m, ST, ED, K, Fish;
int ID[MAXN*10+10], idcnt;
int N;
void Get_E()
{
	memset(E.A, 0x3f, sizeof(E.A));
	for(int i = 0; i < MAXN; i++) 
        E.A[i][i] = 0;
}
void New_mul(Matrix AA, Matrix BB, Matrix &Ans)
{
	Matrix D;
    memset(Ans.A, 0x3f, sizeof(D.A));
    Ans.n = D.n = AA.n;
    Ans.m = D.m = BB.m;
	for(int k = 0; k < MAXN; k++)  
        for(int i = 0; i < MAXN; i++)  
            for(int j = 0; j < MAXN; j++)  
                if(Ans.A[i][j] > AA.A[i][k] + BB.A[k][j])  
                    Ans.A[i][j] = AA.A[i][k] + BB.A[k][j];
}
void pow_mod(Matrix A, int k, Matrix &Ans)
{
    Matrix ans = E, t = A, tmp;
    Ans.n = ans.n = A.n;
    Ans.m = ans.m = A.m;
    while(k)
    {
        if(k & 1)
		{
			New_mul(ans, t, tmp);
			ans = tmp;
		}
        New_mul(t, t, tmp);
		t = tmp;
        k >>= 1;
    }
    memcpy(Ans.A, ans.A, sizeof(ans.A));
}
Matrix Ans;
int main()
{
	ios::sync_with_stdio(false);
	Get_E();
	memset(Map, 0x3f, sizeof(Map));
	cin >> n >> m >> ST >> ED;
	ST--; ED--;
	memset(ID, -1, sizeof(ID));
	for(int i = 0; i < m; i++)
	{
		int u, v, d;
		cin >> d >> u >> v;
		u--;
		v--;
		if(~ID[u] == 0) ID[u] = idcnt++;
		if(~ID[v] == 0) ID[v] = idcnt++;
		Map[ID[u]][ID[v]] = Map[ID[v]][ID[u]] = d;
	}
	E.n = E.m = Z.n = Z.m = A.n = A.m = MAXN;
	memcpy(A.A, Map, sizeof(Map));
	pow_mod(A, n, Ans);
	cout << Ans.A[ID[ST]][ID[ED]] << '\n';
}