随笔 - 375  文章 - 2  评论 - 104  0

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", &n, &m);
41     for (int i = 1; i <= m; ++i){
42         scanf("%d %d %d\n", &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\n", &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\n", x); else printf("%d%c%d\n", x, '/', y);
76 }
View Code

 

posted on 2014-11-03 17:45  Xs酱~  阅读(69)  评论(0编辑  收藏