题意

$n$个点， $m$条边的图，边有费用。

$q$次询问：若每条边的容量为 $\displaystyle\frac uv$，则从 $1$到 $n$流 $1$个单位的流量的最小费用是多少，要求输出最简分数。

$n\le50,m\le100,q\le10^5,0\le u\le v\le10^9$

题解

首先设 ${\rm cost}(x,y)$表示每条边容量为 $y$，要从 $1$向 $n$流 $x$的流量的费用。则有 $\displaystyle{\rm cost}(x,y)=y{\rm cost}(\frac xy,1)$。

根据费用流的算法流程: ${\rm cost}(ny+y,y)-{\rm cost}(ny,y)$表示每条边容量为 $y$时，第 $n+1$条增广路的费用（流量为y）。若只需要 $z$的流量,则乘上比例系数 $\displaystyle\frac zy$即可。

对于 ${\rm cost}(1,p)$，设 $\displaystyle k=\lfloor\frac1p\rfloor$。考虑对前 $k$条增广路每条路流 $p$的流量，费用为 ${\rm cost}(kp,p)=p{\rm cost}(k,1)$；对第 $k+1$条增广路流 $1-kp$的流量，费用为 $\displaystyle\frac{1-kp}{p}[{\rm cost}(kp+p,p)-{\rm cost}(kp,p)]=(1-kp)[{\rm cost}(k+1,1)-{\rm cost}(k,1)]$。

设 $f_k={\rm cost}(k,1)$，则有
${\rm cost}(1,p)=pf_k+(1-pk)(f_{k+1}-f_k)$

综上，设所有边容量为 $1$，做一次最小费用最大流求出所有的 $f_k$。对于一次询问 $\displaystyle p=\frac uv$，则答案为 $\displaystyle\frac{uf_k+(v-ku)(f_{k+1}-f_k)}{v}$，化成最简分数即可。

单组数据时间复杂度 $O(n^2m+q\log V)$

#include <bits/stdc++.h>
using namespace std;
const int N = 50 + 5, M = 2 * 1000 + 5;
const long long Inf = 1e17;
using ll = long long;
using arr = int[N];
/*-----------------------------------------------------------------*/
int n, m, q, Ecnt, Fi[N];
ll dis[N], f[N];
struct Edge {
    int nx, v, c, w;
} E[M];
inline void Add(int u, int v, int c, int w) {
    E[++Ecnt] = (Edge){Fi[u], v, c, w}, Fi[u] = Ecnt;
}
class MCMF {
  private:
    static const int N = 2e5 + 5;
    ll h[N];
    int Flow[N], Pre[N], From[N];
    struct Seg {
        int Cnt, tr[N << 2];
        inline void Cls() {
            fill(tr, tr + 2 * Cnt + 1, 0);
            Cnt = 0;
        }
        inline int Cmp(int a, int b) { return dis[a] < dis[b] ? a : b; }
        inline void Set(int n) {
            for (Cnt = 1; Cnt < n + 2; Cnt <<= 1)
                ;
            tr[0] = n + 1;
        }
        inline void Mdy(int u, ll w) {
            for (int i = u + Cnt; dis[tr[i]] > w; tr[i] = u, i >>= 1)
                ;
            dis[u] = w;
        }
        inline void Del(int u) {
            for (tr[u += Cnt] = 0, u >>= 1; u;
                 tr[u] = Cmp(tr[u << 1], tr[u << 1 | 1]), u >>= 1)
                ;
        }
    } zkw;
    inline bool Dij(int n, int S, int T) {
        for (int i = 0; i <= n; ++i)
            dis[i] = Inf, Flow[i] = From[i] = Pre[i] = 0;
        Flow[S] = 1e9;
        dis[n + 1] = 0;
        zkw.Cls(), zkw.Set(n);
        zkw.Mdy(S, 0);
        for (int _ = 2; _ <= n; ++_) {
            int u = zkw.tr[1];
            zkw.Del(u);
            for (int i = Fi[u], v = E[i].v; i; v = E[i = E[i].nx].v)
                if (E[i].c > 0 && dis[v] > dis[u] + E[i].w + h[u] - h[v]) {
                    zkw.Mdy(v, dis[u] + E[i].w + h[u] - h[v]);
                    From[v] = u, Pre[v] = i;
                    Flow[v] = min(Flow[u], E[i].c);
                }
        }
        return dis[T] != Inf;
    }

  public:
    int MF;
    inline void Calc(int n, int S, int T) {
        int Now;
        for (int i = 1; i <= n; ++i)
            h[i] = 0;
        while (Dij(n, S, T)) {
            MF += (Now = Flow[T]);
            f[MF] = f[MF - Now] + Now * (dis[T] - h[S] + h[T]);
            for (int v = T; v != S; v = From[v])
                E[Pre[v]].c -= Now, E[Pre[v] ^ 1].c += Now;
            for (int i = 1; i <= n; ++i)
                h[i] += dis[i];
        }
    }
} D;
inline void CLS() {
    Ecnt = 1;
    D.MF = 0;
    for (int i = 1; i <= n; ++i)
        Fi[i] = f[i] = 0;
}
inline void Solve() {
    CLS();
    for (int u, v, w; m--;) {
        scanf("%d%d%d", &u, &v, &w);
        Add(u, v, 1, w), Add(v, u, 0, -w);
    }
    D.Calc(n, 1, n);
    scanf("%d", &q);
    for (ll u, v, k; q--;) {
        scanf("%lld%lld", &u, &v);
        if (u == 0 || (k = v / u) > D.MF || (v - u * k > 0 && !f[k + 1]))
            puts("NaN");
        else {
            ll a = (v - k * u) * (f[k + 1] - f[k]) + u * f[k], b = v,
               g = __gcd(a, b);
            printf("%lld/%lld\n", a / g, b / g);
        }
    }
}
int main() {
    while (~scanf("%d%d", &n, &m))
        Solve();
    return 0;
}