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

output:

90

大意：T條已知路徑（雙向），N個地標，輸出從N到1的最短路。

分析：單源最短路問題。

注意：給出的已知路徑中，可能出現重複始末位置的不同長度的路，選取最優的一條儲存。

所以多一個對輸入的路徑是否為該始末位置的最優路徑的判斷：if(e[a][b]>c)   e[b][a]=e[a][b]=c;

//讀入邊
    for(int i=0;i<t;i++)
    {
        cin>>a>>b>>c;
        if(e[a][b]>c)
        e[b][a]=e[a][b]=c;
    }
 

完整程式碼：

#include<iostream>
#include<string>
#include<cstring>
#include<algorithm>
#define inf 0x3fffffff
using namespace std;
int e[1005][1005];
int main()
{
    int t,n,a,b,c,minn,u,dis[1005],book[1005]={0};
    cin>>t>>n;
//e初始化
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
    {
        if(i==j) e[i][j]=0;
        else e[i][j]=inf;
    }
//讀入邊
    for(int i=0;i<t;i++)
    {
        cin>>a>>b>>c;
        if(e[a][b]>c)
        e[b][a]=e[a][b]=c;
    }
//dis初始化
    for(int i=1;i<=n;i++)
        dis[i]=e[1][i];
    book[1]=1;
    for(int i=1;i<n;i++)
    {
//尋找距離1最小的邊
        minn=inf;
        for(int j=1;j<=n;j++)
        {
            if(minn>dis[j]&&book[j]==0)
            {
                minn=dis[j];
                u=j;
            }
        }
        book[u]=1;//標記
        for(int j=1;j<=n;j++)
        {
            if(e[u][j]<inf)
            {
                if(dis[j]>dis[u]+e[u][j])
                   dis[j]=dis[u]+e[u][j];
            }
        }
    }
    cout<<dis[n]<<endl;
    return 0;
}