Bzoj1050 [haoi2006] Travel comf

Handling...

The number of sides is 5000.

So I checked the Internet and found that everyone agreed that the smallest edge of the brute force enumeration was used, and then the query set was used to solve the problem.

The complexity of O (M ^ 2) seems to be able to pass?

Then we started to write brute-force programs. It was too ugly because of a headache.

It's amazing. It's over 7000 Ms. Isn't it a battle without O2? It's over...

 

 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 first18 #define two second19 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

 

Bzoj1050 [haoi2006] Travel comf