给你一个无向图，N(N<=500)个顶点, M(M<=5000)条边，每条边有一个权值Vi(Vi<30000)。给你两个顶点S和T，求一条路径，使得路径上最大边和最小边的比值最小。如果S和T之间没有路径，输出”IMPOSSIBLE”，否则输出这个比值，如果需要，表示成一个既约分数。 备注： 两个顶点之间可能有多条路径。

 #include<bits/stdc++.h>

 #define IOS ios::sync_with_stdio(false);//不可再使用scanf printf

 #define Max(a, b) ((a) > (b) ? (a) : (b))//禁用于函数，会超时

 #define Min(a, b) ((a) < (b) ? (a) : (b))

 #define Mem(a) memset(a, 0, sizeof(a))

 #define Dis(x, y, x1, y1) ((x - x1) * (x - x1) + (y - y1) * (y - y1))

 #define MID(l, r) ((l) + ((r) - (l)) / 2)

 #define lson ((o)<<1)

 #define rson ((o)<<1|1)

 #define Accepted 0

 #pragma comment(linker, "/STACK:102400000,102400000")//栈外挂

 using namespace std;

 inline int read()

 {

     int x=,f=;char ch=getchar();

     while (ch<''||ch>''){if (ch=='-') f=-;ch=getchar();}

     while (ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

     return x*f;

 }

 typedef long long ll;

 const int maxn =  + ;

 const int MOD = ;//const引用更快，宏定义也更快

 const int INF = 1e9 + ;

 const double eps = 1e-;

 struct edge

 {

     int u, v, w;

     edge(){}

     edge(int u, int v, int w):u(u), v(v), w(w){}

     bool operator < (const edge& a)const

     {

         return w < a.w;

     }

 };

 vector<edge>e;

 int p[maxn];

 vector<int>G[maxn];

 void addedge(int u, int v, int w)

 {

     e.push_back(edge(u, v, w));

     G[u].push_back(e.size() - );

     e.push_back(edge(v, u, w));

     G[v].push_back(e.size() - );

 }

 int Find(int x)

 {

     return x == p[x] ? x : p[x] = Find(p[x]);

 }

 struct node

 {

     int a, b;

     node(){}

     node(int A, int B)

     {

         int g = __gcd(A, B);

         a = A / g, b = B / g;

     }

     bool operator < (const node& tmp)const

     {

         return a * tmp.b < b * tmp.a;

     }

 };

 int main()

 {

     int n, m, u, v, w, s, t;

     scanf("%d%d", &n, &m);

     while(m--)

     {

         scanf("%d%d%d", &u, &v, &w);

         addedge(u, v, w);

     }

     scanf("%d%d", &s, &t);

     sort(e.begin(), e.end());

     m = e.size();

     node ans(, );//最大

     for(int start = ; start < m; start++)//枚举下限

     {

         for(int i = ; i <= n; i++)p[i] = i;//初始化并查集

         int cnt = -INF;

         for(int i = start; i < m; i++)

         {

             int u = e[i].u, v = e[i].v, w = e[i].w;

             u = Find(u);

             v = Find(v);

             p[u] = v;

             if(Find(s) == Find(t))

             {

                 cnt = w;

                 break;

             }

         }

         if(cnt == -INF)

         {

             break;

         }

         ans = min(ans, node(cnt, e[start].w));

     }

     if(ans.a ==  && ans.b == )puts("IMPOSSIBLE\n");

     else if(ans.b != )printf("%d/%d\n", ans.a, ans.b);

     else printf("%d\n", ans.a);

     return Accepted;

 }