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
Hint
INPUT DETAILS:
There are five landmarks.
OUTPUT DETAILS:
Bessie can get home by following trails 4, 3, 2, and 1.
解题思路：题目大意就是求最短路,从n到1的最短路。
关于最短路径的思想在前面的博客有；
代码如下：
1 #include<iostream>
2 #include<stdio.h>
3 #include<queue>
4 #include<string.h>
5 using namespace std ;
6
7 const int INF = 0x3f3f3f3f;
8 int G[2000][2000];
9 int d[2000];
10
11 int i ,j;
12
13 struct node{
14 int num;
15 int dis;
16 friend bool operator<(node a ,node b)
17 {
18 return a.dis>b.dis;
19 }
20 };
21
22 int main()
23 {
24 int M , N;
25 int x,y,D;
26
27 priority_queue<node>que;
28
29
30 while(scanf("%d%d",&M,&N)!=EOF)
31 {
32 for( i = 1 ;i <= N ;i++)
33 {
34 for( j = 1 ;j <= N ;j++)
35 {
36 G[i][j] = INF;
37 }
38 }
39
40 for(int i = 1 ; i <= N ;i++)
41 {
42 G[i][i] = 0;
43 }
44 for( i = 1; i <= M ;i++)
45 {
46 scanf("%d%d%d",&x,&y,&D);
47
48 if(G[x][y]>D)
49 {
50 G[x][y] = D;
51 G[y][x] = D;
52 }
53
54 }
55 memset(d,0x3f,sizeof(d));
56 d[1] = 0;
57 que.push({1,0});
58 while(!que.empty())
59 {
60 node tp = que.top();
61
62 que.pop();
63
64 for(i = 1 ;i <= N ;i++)
65 {
66 if(G[tp.num][i])
67 {
68 if(d[i]>d[tp.num]+G[tp.num][i])
69 {
70 d[i] = d[tp.num] + G[tp.num][i];
71 que.push({i,d[i]});
72 }
73 }
74 }
75 }
76
77 printf("%d\n",d[N]);
78 while(!que.empty())
79 {
80 que.pop();
81 }
82
83
84 }
85 return 0;
86 }
原文地址：https://www.cnblogs.com/yewanting/p/10798524.html