RT，看到这个题的时候口胡了一个直接spfa更新的做法， 思路是存当前的最小流量然后直接依题意更新，感觉是假的但是一遍AC，看题解好像没有思路一样的，求助一下做法正确性。

#include <bits/stdc++.h>

using namespace std;
const int N = 2010, INF = 1 << 30;
int h[N], e[N], ne[N], f[N], w[N], idx;
int d[N], vis[N];
int minf[N];

inline int read()
{
    int x = 0, y = 1; char c = getchar();
    while (c < '0' || c > '9') {if (c == '-') y = -1; c = getchar();}
    while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * y;
}

inline void add(int a, int b, int c, int d)
{
    e[idx] = b;
    f[idx] = c;
    w[idx] = d;
    ne[idx] = h[a];   
    h[a] = idx++; 
}

inline void spfa()
{
    memset(d, 0x3f, sizeof d);
    d[1] = 0;
    minf[1] = INF;
    queue<int> q;
    q.push(1);
    while (q.size())
    {
        int u = q.front();
        q.pop();
        vis[u] = false;
        for (int i = h[u]; ~i; i = ne[i])
        {
            int v = e[i];
            if (minf[v] * 1.0 / d[v] < min(minf[u], f[i]) * 1.0 / (d[u] + w[i]))
            {
                minf[v] = min(minf[u], f[i]);
                d[v] = d[u] + w[i];
                if (!vis[v]) 
                    q.push(v), vis[v] = true;
            }
        }
    }
}

int main()
{
    memset(h, -1, sizeof h);
    int n = read(), m = read();
    for (int i = 1; i <= m; i++)
    {
        int a = read(), b = read(), c = read(), d = read();
        add(a, b, d, c);
        add(b, a, d, c);
    }
    spfa();
    double k = 1000000 * (minf[n] * 1.0 / d[n]);
    printf("%d\n", (int)k);
    return 0;
}