POJ 2387

Til the Cows Come Home

Description

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

思路：spfa

 1 #include<queue>

 2 #include<cstdio>

 3 #include<string>

 4 #include<cstring>

 5 #include<iostream>

 6 #include<algorithm>

 7 #define MAXN 2222

 8 #define INF 0x7fffffff

 9 using namespace std;

10 typedef struct{

11     int to, next, w;

12 }Edge;

13 queue<int>Q;

14 Edge edge[2*MAXN];

15 int head[MAXN/2], vis[MAXN/2], dist[MAXN/2], N, T;

16 void init(){

17     memset(vis, 0, sizeof(vis));

18     for(int i = 1;i < N;i ++) dist[i] = INF;

19     dist[N] = 0;

20 }

21 void addedge(int u, int v, int w, int k){

22     edge[k].to = v;

23     edge[k].w = w;

24     edge[k].next = head[u];

25     head[u] = k++;

26     edge[k].to = u;

27     edge[k].w = w;

28     edge[k].next = head[v];

29     head[v]=k;

30 }

31 void spfa(int s){

32     init();

33     while(!Q.empty()) Q.pop();

34     vis[s] = 1;

35     Q.push(s);

36     while(!Q.empty()){

37         int p = Q.front();

38         Q.pop();

39         vis[p] = 0;

40         for(int i = head[p];~i;i = edge[i].next){

41             int u = edge[i].to;

42             if(dist[u] > dist[p] + edge[i].w){

43                 dist[u] = dist[p] + edge[i].w;

44                 if(!vis[u]){

45                     Q.push(u);

46                     vis[u] = 1;

47                 }

48             }

49         }

50     }

51 }

52 int main(){

53     int u, v, w;

54     /* freopen("in.c", "r", stdin); */

55     while(~scanf("%d%d", &T, &N)){

56         int k = 0;

57         memset(head, -1, sizeof(head));

58         for(int i = 0;i < T;i ++){

59             scanf("%d%d%d", &u, &v, &w);

60             addedge(u, v, w, k);

61             k += 2;

62         }

63         spfa(N);

64         printf("%d\n", dist[1]);

65     }

66     return 0;

67 }