最短路Dijkstra——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

output:

90

大意：T条已知路径（双向），N个地标，输出从N到1的最短路。

分析：单源最短路问题。

注意：给出的已知路径中，可能出现重复始末位置的不同长度的路，选取最优的一条保存。

所以多一个对输入的路径是否为该始末位置的最优路径的判断：if(e[a][b]>c)   e[b][a]=e[a][b]=c;

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

完整代码：

#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;
    }
//读入边
    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++)
    {
//寻找距离1最小的边
        minn=inf;
        for(int j=1;j<=n;j++)
        {
            if(minn>dis[j]&&book[j]==0)
            {
                minn=dis[j];
                u=j;
            }
        }
        book[u]=1;//标记
        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;
}