题意

n个点m条边的有向图，给出边的单位流量费用，q次询问，每次询问有有u,v两个整数，表示所有边的容量为u/v，对于每次询问，输出从1到n流量为1时的最小费用，流量达不到1时，输出“NaN”.

题解

每条边的容量是相同的，如果边的容量为1的时候流量为maxflow，那么边的容量为u/v时，流量就是u* maxflow/v。
所以，只要将所有边按容量为1跑最小费用最大流，将每次增广的流量和费用存下来，按费用从小到大排序，对于每次询问，假定边容量为u，那么最大流就是u*maxflow，按费用从小到大加到将流量凑到v，得到的费用再除以v，就是流量为1时的费用了。

代码

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
const int N = 5e3 + 7;
const int M = 1e4 + 7;
const int maxn = 5e3 + 7;
typedef long long ll;
int maxflow, mincost;
struct Edge {
    int from, to, cap, flow, cost;
    Edge(int u, int v, int ca, int f, int co):from(u), to(v), cap(ca), flow(f), cost(co){};
};
struct MCMF
{
    int n, m, s, t;
    vector<Edge> edges;
    vector<int> G[N];
    int inq[N], d[N], p[N], a[N];//是否在队列 距离 上一条弧 可改进量
    void init(int n) {
        this->n = n;
        for (int i = 0; i < n; i++) G[i].clear();
        edges.clear();
    }
    void add(int from, int to, int cap, int cost) {
        edges.push_back(Edge(from, to, cap, 0, cost));
        edges.push_back(Edge(to, from, 0, 0, -cost));
        int m = edges.size();
        G[from].push_back(m - 2);
        G[to].push_back(m - 1);
    }
    vector<pair<int, int> > v;
    bool SPFA(int s, int t, int &flow, int &cost) {
        for (int i = 0; i < N; i++) d[i] = INF;
        memset(inq, 0, sizeof(inq));
        d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF;
        queue<int> que;
        que.push(s);
        while (!que.empty()) {
            int u = que.front();
            que.pop();
            inq[u]--;
            for (int i = 0; i < G[u].size(); i++) {
                Edge& e = edges[G[u][i]];
                if(e.cap > e.flow && d[e.to] > d[u] + e.cost) {
                    d[e.to] = d[u] + e.cost;
                    p[e.to] = G[u][i];
                    a[e.to] = min(a[u], e.cap - e.flow);
                    if(!inq[e.to]) {
                        inq[e.to]++;
                        que.push(e.to);
                    }
                }
            }
        }
        if(d[t] == INF) return false;
        flow += a[t];
        cost += d[t] * a[t];
        v.push_back(make_pair(d[t], a[t]));
        int u = t;
        while (u != s) {
            edges[p[u]].flow += a[t];
            edges[p[u]^1].flow -= a[t];
            u = edges[p[u]].from;
        }
        return true;
    }
    int MinMaxflow(int s, int t) {
        int flow = 0, cost = 0;
        while (SPFA(s, t, flow, cost));
        maxflow = flow; mincost = cost;
        return cost;
    }
};
int main()
{
    int n, m, s, t;
    while (scanf("%d%d", &n, &m) != EOF) {
        MCMF solve;
        s = 1, t = n;
        for (int i = 1, a, b, c; i <= m; i++) {
            scanf("%d%d%d", &a, &b, &c);
            solve.add(a, b, 1, c);
        }
        solve.MinMaxflow(s, t);
        sort(solve.v.begin(), solve.v.end());
        int q; ll u, v; scanf("%d", &q);
        while (q--) {
            scanf("%lld%lld", &u, &v);
            if(u * maxflow < v) printf("NaN\n");
            else {
                ll sumc = 0, sumf = 0;
                for (int i = 0; i < solve.v.size(); i++) {
                    int flow = solve.v[i].second, cost = solve.v[i].first;
                    if(sumf + u <= v) {
                        sumf += u;
                        sumc += u * cost;
                    }
                    else if(sumf < v) {
                        sumc += (v - sumf) * cost;
                        sumf = v;
                    }
                    else break;
                }
                if(sumf != v) printf("NaN\n");
                else {
                    ll gcd = __gcd(sumc, v);
                    printf("%lld/%lld\n", sumc/gcd, v/gcd);
                }
            }
        }
    }
}