POJ 2387 Til the Cows Come Home(Dijkstra判重邊)


Til the Cows Come Home
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 31748   Accepted: 10757

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.

Source


題目大意:

說的是,一只奶牛位於N號節點,輸入N個節點和T對雙向的邊,求出由N到1的最短的距離,其實就是問的單源最短路問題,首先想到了Dijkstra算法。


解題思路:

直接水過去吧,但是有一點非常重要,我也是因為這點給WA了一次,那就是對於重邊的判斷了,如果在雙向邊中,我們當然要取的是那個最短的那一次了,,因為是求最短路徑了。代碼我寫的很規范了,,也是很容易理解的。。

代碼:

# include<cstdio>
# include<iostream>

using namespace std;

# define MAX 1000+4
# define inf 99999999

int n,t;
int edge[MAX][MAX];
int dis[MAX];
int book[MAX];
int u,v;

void init()
{
    for ( int i = 1;i <= n;i++ )
    {
        for ( int j = 1;j <= n;j++ )
        {
            if ( i==j )
            {
                edge[i][j] = 0;
            }
            else
            {
                edge[i][j] = inf;
            }
        }
    }
}

void input()
{
    for ( int i = 0;i < t;i++ )
    {
        int t1,t2,t3;
        cin>>t1>>t2>>t3;
        if ( t3 < edge[t1][t2] )
        {
            edge[t1][t2] = t3;
            edge[t2][t1] = t3;
        }
    }

}


void Dijkstra()
{
    for ( int i = 1;i <= n;i++ )
    {
        book[i] = 0;
        dis[i] = edge[n][i];
    }

    int _min;
    for ( int i = 1;i <= n-1;i++ )
    {
       _min = inf;
       for ( int j = 1;j <= n;j++ )
       {
           if ( book[j]==0&&dis[j] < _min )
           {
               _min = dis[j];
                  u = j;
           }
       }

       book[u] = 1;
       for ( v = 1;v <= n;v++ )
       {
           if ( book[v]==0&&dis[v] > dis[u]+edge[u][v] )
           {
               dis[v] = dis[u]+edge[u][v];
           }
       }

    }
}

int main(void)
{
    while ( cin>>t>>n )
    {
        init();
        input();
        Dijkstra();
        cout<<dis[1]<<endl;

    }


    return 0;
}



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