给出n种物品,这些物品可以进行转换。给出m条转换规则,形如: a a a 个 b b b 物品 可以转换为 c × w c \times w c×w 个 d d d 物品。问: w w w 最大取多少,才能使这些物品无法转换成无限多的物品。
明显就是一道图论题目。边: c → d c \to d c→d, 边权: c w a \frac{cw}{a} acw
const int N = 4e3 + 3;
int n, m;
int h[N], net[N], e[N], a[N], c[N], tot;
void add(int x, int y, int z, int zz)
{
e[++tot] = y;
a[tot] = z; c[tot] = zz;
net[tot] = h[x];
h[x] = tot;
}
double d[N];
int v[N], vis[N], cnt[N];
//v标记该点是否在队列中
//vis标记该点是否已经走过,用来遍历森林结构
//cnt标记该点的跟新次数,更新次数 >= n 则存在负环
bool spfa(double w)
{
for (int i = 1; i <= n; i++)
d[i] = 1e18, v[i] = 0, vis[i] = 0;
for (int i = 1; i <= n; i++)
{
if (vis[i]) continue;
vis[i] = 1;
queue<int> q;
q.push(i);
d[i] = 0; v[i] = 1;
cnt[i] = 0;
while (sz(q))
{
int x = q.front(); q.pop();
v[x] = 0;
for (int i = h[x]; i; i = net[i])
{
int y = e[i]; double z = -log(w * c[i] / a[i]);
if (d[y] > d[x] + z)
{
vis[x] = 1;//标记走过的点
cnt[y] = cnt[x] + 1;
if (cnt[y] >= n) return 0;//发现负环
d[y] = d[x] + z;
if (v[y] == 0) q.push(y), v[y] = 1;
}
}
}
}
return 1;
}
int main()
{
IOS;
cin >> n >> m;
for (int i = 1, a, b, c, d; i <= m; i++)
{
cin >> a >> b >> c >> d;
add(b, d, a, c);
}
double l = 0, r = 1;
while (r - l > esp)
{
double mid = (l + r) / 2;
if (spfa(mid)) l = mid;
else r = mid;
}
cout << fixed << setprecision(10) << r << endl;
return 0;
}