題目描述：

給定的無向圖有n個頂點，m條邊；求出從頂點1到頂點m的最短路徑； 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.

思路

Prim演算法： 裸的Dijkstra演算法資料範圍小，不需要優化記憶體及演算法就可以ac

程式碼:

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
 


#define MAX 2010
#define INF 0x3f3f3f3f

int mat[MAX][MAX], vis[MAX], dist[MAX];

void init(int n)
{
    memset(vis, 0, sizeof(vis));
    for(int i = 1; i<=n; i++)
    {
        for(int j = 1; j<=n; j++)
            mat[i][j] = INF;
        dist[i] = INF;
    }
}

void Union(int x, int y, int val)
 

{
    if(val < mat[x][y])
        mat[x][y] = mat[y][x] = val;
}

void Dijkstra(int n)
{
    dist[1] = 0;
    for(int i = 0; i<n; i++)
    {
        int min_vertex, min_dist = INF;
        for(int j = 1; j<=n; j++)
        {
            if(!vis[j] && dist[j] < min_dist)
            {
                min_dist = dist[j];
                min_vertex = j;
            }
        }
        vis[min_vertex] = 1;
        for(int j = 1; j<=n; j++)
        {
            if(!vis[j] && mat[min_vertex][j] + min_dist < dist[j])
            {
                dist[j] = mat[min_vertex][j] + min_dist;
            }
        }
    }
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n, m;
    while(cin >> n >> m)
    {
        init(m);
        for(int i = 0; i<n; i++)
        {
            int x, y, val;
            cin >> x >> y >> val;
            Union(x, y, val);
        }
        Dijkstra(m);
        cout << dist[m] << endl;
    }
    return 0;
}