搬运。。。
一看题，边数5000，百思不得其解。
于是上网查，发现大家一致说暴力枚举最小边，然后并查集求解。
O(M ^ 2)的复杂度，好像能过?
然后就开始写暴力程序，因为头疼，写的太难看了。
真是神奇，7000+Ms还算过了，是不是不开O2就会TLE呢?反正过了。。。
1 /************************************************************** 2 Problem: 1050 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:7548 ms 7 Memory:1396 kb 8 ****************************************************************/ 9 10 #include <cstdlib> 11 #include <cstdio> 12 #include <cmath> 13 #include <algorithm> 14 #include <iostream> 15 #include <utility> 16 17 #define one first 18 #define two second 19 using namespace std; 20 21 pair <int, pair<int, int> > a[10000]; 22 int m, n, s, t, f1, f2, fa[1000]; 23 24 int find(int x){ 25 int f = fa[x]; 26 if (f == x) return f; 27 f = find(f); 28 fa[x] = f; 29 return f; 30 } 31 32 void add(int x, int y){ 33 int f1 = find(x), f2 = find(y); 34 if (f1 != f2) fa[f1] = f2; 35 } 36 37 int main(){ 38 int x, y, v, anss = 0; 39 double ans = 100000000; 40 scanf("%d %d ", &n, &m); 41 for (int i = 1; i <= m; ++i){ 42 scanf("%d %d %d ", &x, &y, &v); 43 a[i].one = v; 44 a[i].two.one = x; 45 a[i].two.two = y; 46 } 47 scanf("%d %d ", &s, &t); 48 sort(a + 1, a + m + 1); 49 for (int i = 1; i <= m; ++i){ 50 for (int j = 1; j <= n; ++j) 51 fa[j] = j; 52 for (int j = i; j <= m; ++j){ 53 add(a[j].two.one, a[j].two.two); 54 f1 = find(s); 55 f2 = find(t); 56 if (f1 == f2) 57 if (ans > (double)a[j].one / a[i].one){ 58 ans = (double)a[j].one / a[i].one; 59 anss = i; 60 x = a[j].one; 61 y = a[i].one; 62 break; 63 } 64 } 65 } 66 if (anss == 0){ 67 cout<<"IMPOSSIBLE"<<endl; 68 return 0; 69 } 70 for (int i = 2; i <= y; ++i) 71 while (x % i == 0 && y % i == 0){ 72 x /= i; 73 y /= i; 74 } 75 if (y == 1) printf("%d ", x); else printf("%d%c%d ", x, '/', y); 76 }