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
Hint
INPUT DETAILS:There are five landmarks.
OUTPUT DETAILS:
Bessie can get home by following trails 4, 3, 2, and 1. 题意：有n个点，给出几组数据（两点 和两点之间的距离），求从1 到其他点之间的最短距离。输入：第一行，t,n.t表示给出的数据，n表示共有几个点，接下来t行每行为两个点和两点之间的距离，。输出从1到n之间的最短距离。 思路：最短路问题，题中无负权，所以可以用Floyd算法和Dijkstra算法，但是数据要求为1000，比较大，因此用Dijkstra算法。 代码
1 #include <cstdio>
2 #include <fstream>
3 #include <algorithm>
4 #include <cmath>
5 #include <deque>
6 #include <vector>
7 #include <queue>
8 #include <string>
9 #include <cstring>
10 #include <map>
11 #include <stack>
12 #include <set>
13 #include <sstream>
14 #include <iostream>
15 #define mod 998244353
16 #define eps 1e-6
17 #define ll long long
18 #define INF 0x3f3f3f3f
19 using namespace std;
20
21 int t,n;
22 //ma[i][j]存放i到j点之间的距离，dis数组存放1到各点之间的距离
23 int ma[1010][1010],dis[1010];
24 void dijkstra(int m)
25 {
26 //vis数组标记已经找过的最短路
27 bool vis[1010];
28 memset(vis,0,sizeof(vis));
29 vis[1]=1;
30 for(int i=1;i<m;i++)
31 {
32 //mi记录当前到1距离最小的值
33 int mi=INF;
34 //k表示到1距离最小的点
35 int k=1;
36 for(int j=1;j<=m;j++)
37 {
38 //当未标记这个点，并且这个点到1的距离时最小的时侯成立，
39 if(!vis[j]&&mi>dis[j])
40 {
41 mi=dis[j];
42 k=j;
43 }
44 }
45 //已找到1到k点的最小值，所以标记这个点
46 vis[k]=1;
47 for(int j=1;j<=m;j++)
48 {
49 //当未标记这个点，且j点到1的距离比1到k的距离加上k到j点的距离小时，更换dis的数据
50 if(!vis[j]&&dis[j]>(dis[k]+ma[k][j]))
51 {
52 dis[j]=dis[k]+ma[k][j];
53 }
54 }
55 }
56 }
57 int main()
58 {
59 scanf("%d %d",&t,&n);
60 //初始化ma数组
61 for(int i=1;i<=n;i++)
62 {
63 //i点和i点之间距离为0
64 ma[i][i]=0;
65 for(int j=1;j<i;j++)
66 {
67 //初始两点之间距离为INF,
68 ma[i][j]=ma[j][i]=INF;
69 }
70 }
71 int a,b,c;
72 for(int i=1;i<=t;i++)
73 {
74 scanf("%d %d %d",&a,&b,&c);
75 //如果输入的两点有重复，则保存小的
76 if(c<ma[a][b])
77 {
78 ma[a][b]=ma[b][a]=c;
79 }
80 }
81 for(int i=1;i<=n;i++)
82 {
83 //dis数组保存从1到其他点之间的距离
84 dis[i]=ma[1][i];
85 }
86 dijkstra(n);
87 printf("%d\n",dis[n]);
88 }