Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible.
Farmer John‘s field has N (2 <= N <= 1000) landmarks in
it, uniquely numbered 1..N. Landmark 1 is the barn; the apple tree
grove in which Bessie stands all day is landmark N. Cows travel in the
field using T (1 <= T <= 2000) bidirectional cow-trails of various
lengths between the landmarks. Bessie is not confident of her
navigation ability, so she always stays on a trail from its start to its
end once she starts it.
Given the trails between the landmarks, determine the minimum
distance Bessie must walk to get back to the barn. It is guaranteed
that some such route exists.
Input
* Line 1: Two integers: T and N
* Lines 2..T+1: Each line describes a trail as three
space-separated integers. The first two integers are the landmarks
between which the trail travels. The third integer is the length of the
trail, range 1..100.
Output
* Line 1: A single integer, the minimum distance that Bessie must travel to get from landmark N to landmark 1.
Sample Input
5 5 1 2 20 2 3 30 3 4 20 4 5 20 1 5 100
Sample Output
90 O（n^2)
1 #include <cstdio>
2 #include <cstring>
3 #include <iostream>
4 #include <algorithm>
5 #include <vector>
6 #include <queue>
7 #include <set>
8 #include <map>
9 #include <string>
10 #include <cmath>
11 #include <stdlib.h>
12 #define MAXSIZE 1005
13 using namespace std;
14
15 int t,n;
16 int m[MAXSIZE][MAXSIZE],dis[MAXSIZE],vis[MAXSIZE];
17
18 void dij(int s)
19 {
20 int i,j,k;
21 for(i = 0;i<n;i++)
22 {
23 dis[i] = m[s][i];
24 vis[i] = 0;
25 }
26 dis[s] = 0;
27 vis[s] = 1;
28 for(i = 1;i<=n;i++)
29 {
30 k = 0;
31 int mn = 1000000;
32 for(j = 1;j<=n;j++)
33 if(!vis[j]&&dis[j]<mn)
34 {
35 mn = dis[j];
36 k = j;
37 }
38 vis[k] = 1;
39 for(j = 1;j<=n;j++)
40 if(!vis[j]&&dis[j]>dis[k]+m[k][j])
41 dis[j] = dis[k]+m[k][j];
42 }
43 }
44
45 int main()
46 {
47 freopen("caicai.txt","r",stdin);
48 cin>>t>>n;
49 int i,j,a,b,w;
50 for(i = 0;i<=n;i++)
51 for(j = 0;j<=n;j++)
52 m[i][j] = 1000;
53 for(i = 0;i<t;i++)
54 {
55 scanf("%d%d%d",&a,&b,&w);
56 if(m[a][b]>w)
57 {
58 m[a][b] = w;
59 m[b][a] = w;
60 }
61 }
62 dij(n);
63 cout<<dis[1]<<endl;
64 return 0;
65 }