Til the Cows Come Home (dijkstra)

题目

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.

翻译

贝西在田里，想在农夫约翰叫醒她早上挤奶之前回到谷仓尽可能多地睡一觉。贝西需要她的美梦，所以她想尽快回来。

农场主约翰的田里有n（2<=n<=1000）个地标，唯一编号为1…n。地标1是谷仓；贝西整天站在其中的苹果树林是地标n。奶牛在田里行走时使用地标间不同长度的T（1<=t<=2000）双向牛道。贝西对自己的导航能力没有信心，所以一旦开始，她总是沿着一条从开始到结束的路线行进。

根据地标之间的轨迹，确定贝西返回谷仓必须走的最小距离。这样的路线一定存在。
输入
*第1行：两个整数：t和n

*第2.t+1行：每行将一条轨迹描述为三个空格分隔的整数。前两个整数是两条轨迹之间的标志。第三个整数是轨迹的长度，范围为1…100。
输出
*第1行：一个整数，贝西从N地标到1地标所需的最小距离。
样本输入
5 5
1 2 20
2 3 30
3 4 20
4 5 20
15 100
样品输出
90
提示
输入详细信息：

有五个标志。

输出详细信息：

贝西可以沿着小路4、3、2和1回家。

思路

输入图的边信息.起点 终点 权值.求第一个点到最后一个点的最短距离.
直接用dijkstra，也可以用stl优先队列

#include<stdio.h>
#define Max 0x3fffffff
int map[1005][1005];
int dis[1005];

void dijkstra(int n)
{
    int visit[1001]={0};
    int min,i,j,k;
    visit[1]=1;
    for(i=1;i<n;++i)
    {
        min=Max;
        k=1;
        for(j=1;j<=n;++j)
        {
            if(!visit[j]&&min>dis[j])
            {
                min=dis[j];
                k=j;
            }
        }
        visit[k]=1;
        for(j=1;j<=n;++j)
        {
            if(!visit[j]&&dis[j]>dis[k]+map[k][j])
                dis[j]=dis[k]+map[k][j];
        }
    }
    printf("%d\n",dis[n]);
}

int main()
{
    int t,n,i,j,from,to,cost;
    while(scanf("%d%d",&t,&n)!=EOF)
    {
        for(i=1;i<=n;++i)
        {
            map[i][i]=0;
            for(j=1;j<i;++j)
                map[i][j]=map[j][i]=Max;
        }
        for(i=1;i<=t;++i)
        {
            scanf("%d%d%d",&from,&to,&cost);
            if(cost<map[from][to])                //可能有多条路，只记录最短的
                map[from][to]=map[to][from]=cost;
        }
        for(i=1;i<=n;++i)
            dis[i]=map[1][i];
        dijkstra(n);
    }
    return 0;
}