Til the Cows Come Home

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.

 

题目大意 ：求最短路径；

 

题目分析 ：这道题其实套用Dijkstra算法的模板，一下就可以写完了，属于水题一个。但我还是错了很多次。注意点：先输入边，在输入点。我就是错在这，烦！

 

题目收获 ：看清题意。

 

AC代码 ：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>

using namespace std;

const int MAX = 999999;
const int maxn = 5000;
int G[maxn][maxn];//邻接表;
int T, N;

void Dijkstra(int start)
{
    bool vis[maxn];
    int dis[maxn];
    memset(vis, false, sizeof(vis));
    for (int i = 1; i <= N; i++)
        dis[i] = G[start][i];
    dis[start] = 0;
    vis[start] = true;
    for (int i = 1; i <= N - 1; i++)
    {
        int mi = MAX, x;
        for (int k = 1; k <= N; k++)
            if (!vis[k] && dis[k] < mi)
            {
                mi = dis[k];
                x = k;
                //cout << "X: " << x << "
";
            }
        if (x == start)
            break;
        vis[x] = true;
        for (int k = 1; k <= N; k++)
            if (!vis[k] && dis[k] > dis[x] + G[x][k])
            {
                //cout << "*****************XXXX:" << i << "
";
                //printf("dis[%d]:%d----G[%d][%d]:%d
", x, dis[x], x, k, G[x][k]);
                dis[k] = dis[x] + G[x][k];
                //printf("dis[%d]:%d
", k, dis[k]);
                
            }
    }
    printf("%d
",dis[N]);
}

void init()
{
    int m = T;
    for (int i = 1; i <= N; i++)
        for (int j = 1; j <= N; j++)
            G[i][j] = (i == j ? 0 : MAX);
    while (m--)
    {
        int a, b, c;
        cin >> a >> b >> c;
        if (G[a][b] > c)
            G[a][b] = G[b][a] = c;
    }
}

int main()
{
    cin >> T >> N;
    init();
    Dijkstra(1);
    return 0;
}