Til the Cows Come Home POJ

Til the Cows Come Home

 POJ - 2387 

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 51 2 202 3 303 4 204 5 201 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：求从某点（本题为点1）到点n的最短路径

思路：迪杰斯特拉算法。

注意：本题需要考虑重边。

AC代码：

#include<iostream>#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>#define INF 0x7fffffff#define maxn 2010using namespace std;int Map[maxn][maxn];int vis[maxn];int dis[maxn];int n,m;void Dijs(){int i,j;memset(dis,INF,sizeof(dis));dis[1]=0;for(i=2;i<=n;++i){dis[i]=Map[1][i];}memset(vis,0,sizeof(vis));vis[1]=1;int v,Min;for(i=1;i<=n;++i){                 //找出从前一点出发走最短路径能到达的点  Min=INF;for(j=1;j<=n;++j){if(!vis[j]&&dis[j]<Min){v=j;Min=dis[j];}}vis[v]=1;for(j=1;j<=n;++j){             //比较出从起点到达点j的最短路径if(!vis[j]&&Map[v][j]<INF){dis[j]=min(dis[j],dis[v]+Map[v][j]);}}}printf("%d\n",dis[n]);}int main(){int a,b,c;int i,j;while(scanf("%d%d",&m,&n)!=EOF){for(i=0;i<=n;++i){for(j=0;j<=n;++j){if(i==j)Map[i][j]=0;Map[i][j]=INF;}}for(i=1;i<=m;++i){scanf("%d%d%d",&a,&b,&c);if(Map[a][b]>c){     //当出现重边时，留取最短的一条 Map[a][b]=c;Map[b][a]=c;}}Dijs();}return 0;}