[NWPU][2014][TRN][18]最短路问题 A - 模板 POJ 2387

题意：给出两个整数T,N，然后输入一些点直接的距离，求N和1之间的最短距离。。

思路：dijkstra求单源最短路，但是要注意判重。

A - 模板

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  POJ 2387

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

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>
using namespace std;
#define inf  0xffffff
const int MAXN = 1005;
int maze[MAXN][MAXN], vis[MAXN], mincost[MAXN];
//mincost[i]记录的已经标记的节点（s1）到其他节点i（s2）的权值的最小值
int n, t;

int dijkstra(int v)
{

    for(int i = 1; i <= n; i++)
        mincost[i] = maze[v][i];//以v节点初始化
    vis[v] = 1;//把v标记，记录到集合s1中
    int sum =0, k;
    for(int i = 2; i <= n; i++)//寻找n -1次
    {
        //sum = 0;
        int mina = inf;
        for(int j = 1; j <= n; j++)
        {
            if(mincost[j] < mina && !vis[j])
            {//找到集合s2中最小的权值累加并加入到集合s1中
                mina = mincost[j]; k = j;
            }
        }
        sum += mina;vis[k] = 1;
    //cout << sum << endl;
        for(int j = 1; j <= n; j++)//因为S1集合加入了节点k，所以要更新lowcost
                        //lowcost更新为已经找到的节点到集合S2中的距离（map[k][j]），和原来S1集合
                        //（未加入K）时候到集合S2中的距离的较小值,这个值才是当前S1集合到S2集合
                        //中距离最小的点。
        {
            if(mincost[j] > maze[k][j] + mina && !vis[j])
            {
                mincost[j] = maze[k][j] + mina;
            }
        }
    }
    return mincost[n];
}

int main()
{
    //freopen("input.txt", "r", stdin);
    cin >> t >> n;
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= n; j++)
        {
            //if(i == j)
               // maze[i][i] = 0;
            maze[i][j] = maze[j][i] = inf;
        }
    }
    for(int i = 1; i <= t; i++)
    {
        int a, b, c;
        cin >> a >> b >> c;
        if(maze[a][b] > c)
            maze[a][b] = maze[b][a] = c;
    }
    cout << dijkstra(1) << endl;
    return 0;
}