1 program ddk;
2 type ty=record
3 x,y,z:longint;
4 end;
5 var a:array[1..500]of longint;
6 b:array[0..5000]of ty;
7 n,i,j,m,max,max1,min,min1,a1,a2,s,t:longint;
8 procedure kuai(s,t:longint);
9 var i,j,p:longint;
10 begin
11 i:=s;
12 j:=t;
13 p:=b[(s+t)div 2].z;
14 repeat
15 while b[i].z<p do
16 inc(i);
17 while b[j].z>p do
18 dec(j);
19 if j>=i
20 then
21 begin
22 b[0]:=b[i];
23 b[i]:=b[j];
24 b[j]:=b[0];
25 inc(i);
26 dec(j);
27 end;
28 until i>j;
29 if i<t
30 then kuai(i,t);
31 if j>s
32 then kuai(s,j);
33 end;
34 function zhao(a1:longint):longint;
35 begin
36 if a[a1]<>a1
37 then
38 begin
39 a[a1]:=zhao(a[a1]);
40 exit(a[a1]);
41 end
42 else exit(a[a1]);
43 end;
44 function gcd(a1,a2:longint):longint;
45 var a3:longint;
46 begin
47 repeat
48 a3:=a1 mod a2;
49 a1:=a2;
50 a2:=a3;
51 until a2=0;
52 exit(a1);
53 end;
54 begin
55 max:=9999999;
56 min:=1;
57 read(n,m);
58 for i:=1 to m do
59 read(b[i].x,b[i].y,b[i].z);
60 kuai(1,m);
61 read(s,t);
62 for i:=1 to m do
63 begin
64 for j:=1 to n do
65 a[j]:=j;
66 max1:=b[i].z;
67 min1:=0;
68 for j:=i downto 1 do
69 begin
70 a1:=zhao(b[j].x);
71 a2:=zhao(b[j].y);
72 a[a2]:=a1;
73 if zhao(s)=zhao(t)
74 then
75 begin
76 min1:=b[j].z;
77 break;
78 end;
79 end;
80 if (min1<>0)and(max/min>max1/min1)
81 then
82 begin
83 max:=max1;
84 min:=min1;
85 end;
86 end;
87 if max=9999999
88 then writeln('IMPOSSIBLE')
89 else
90 if max mod min=0
91 then writeln(max div min)
92 else
93 begin
94 a1:=gcd(max,min);
95 writeln(max div a1,'/',min div a1);
96 end;
97 end.
将边按权值排序之后，从小到大枚举最大边，用并查集维护s，t是否相连，刚相连是跳出，算出比值，不断更新答案
