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POJ 2387

时间:2014-05-01 15:42:53      阅读:419      评论:0      收藏:0      [点我收藏+]

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Til the Cows Come Home
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 27591   Accepted: 9303

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

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 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 }
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POJ 2387,布布扣,bubuko.com

POJ 2387

标签:des   style   blog   class   code   java   tar   ext   javascript   width   color   

原文地址:http://www.cnblogs.com/anhuizhiye/p/3700726.html

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