Til the Cows Come Home

Think:
单源最短路问题采用Dijkstra算法TAT。 套模板就好了。。。。
整整写了一个小时感觉没救了， 10个WA， 1个PE 呵呵。

题目大意：
有N个点，给出从a点到b点的距离，当然a和b是互相可以抵达的，问从1到n的最短距离

注意点：
判断重边

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.

#include<iostream>#include<cstdio>#include<string.h>#define INF 0x3f3f3f3f#include<math.h>#include<algorithm>#include<stdlib.h>#include<stdio.h>8using namespace std;bool vis[10086];int Map[3030][3030];int dis[3030];int Dijkstra();int N, T;int main(){    int i, j;    int a, b, c;    cin >> T >> N;    memset(vis, 0, sizeof(vis));    memset(dis, 0, sizeof(dis));    for(i = 1; i <= N; i ++)    {        for (j = 1; j <= N; j ++)        {            Map[i][j] = INF;        }        Map[i][i] = 0;    }    for (i = 1; i <= N; i ++)        dis[i] = INF;    dis[N] = 0;    while(T --)    {        cin >> a >> b >> c;        if (Map[a][b] > c)            Map[a][b] = Map[b][a] = c;    }    int ans = Dijkstra();    cout << ans << endl;}int Dijkstra(){    int i, j, k;    for (i = 1; i <= N; i ++)    {        int u, MIN = INF;        for (j = 1; j <= N; j ++)        {            if (!vis[j] && dis[j] <= MIN)            {                u = j;                MIN = dis[j];            }        }        vis[u] = 1;        for(k = 1; k <= N; k ++)        {            dis[k] = min(dis[k], dis[u] + Map[u][k]);        }    }    return dis[1];}