rt，用了 Floyd+log 求最短路，好像不是精度的问题，但是只有60分。

https://www.luogu.com.cn/record/36542797

#include <cstdio>
#include <cmath>
using namespace std;

typedef long double db;
const int max_n = 200, INF = 0x3f3f3f3f;
const db EPS = 1e-14;

struct costs
{
	int del;
	db los;

	costs(int _d = 0, db _l = 0) : del(_d), los(_l) { }

	bool operator<(const costs& a) const
	{
		if (fabs(los - a.los) > EPS)
			return los < a.los;
		return del < a.del;
	}

	costs operator+(const costs& a) const
	{
		return costs(del + a.del, los + a.los);
	}
};

costs g[max_n][max_n];

int main()
{
	int n, s, t, ta;
	db tb;

	scanf("%d%d%d", &n, &s, &t);
	
	for (int i = 0; i < n; i++)
		for (int j = 0; j < n; j++)
		{
			scanf("%d", &ta);

			if (ta == -1)
				g[i][j].del = INF;
			else
				g[i][j].del = ta;
		}
	
	for (int i = 0; i < n; i++)
		for (int j = 0; j < n; j++)
		{
			scanf("%Lf", &tb);

			if (tb == -1)
				g[i][j].los = 10 * INF;
			else
				g[i][j].los = -log(1 - tb);
		}
	
	for (int k = 0; k < n; k++)
		for (int i = 0; i < n; i++)
			for (int j = 0; j < n; j++)
				if (g[i][k] + g[k][j] < g[i][j])
					g[i][j] = g[i][k] + g[k][j];

	printf("%d %.4Lf\n", g[s-1][t-1].del, 1 - exp(-g[s-1][t-1].los));

	return 0;
}