POJ - 2387 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
  Sample Output
  90
  Hint
  INPUT DETAILS:

There are five landmarks.

OUTPUT DETAILS:
求从1到n的最短路长度
用dijkstra直接求就可以了

#include <iostream>
#include <cstdio>
#include <cstring>
#define inf 0x3f3f3f3f

using namespace std;
int d[1001];//记录起点到各个点的最短路径长度 
int w[1001][1001];//记录有权重的图
bool vis[1001];//访问标记
int t,n;

void dijkstra()
{
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=n;i++)
    {
        d[i]=w[1][i];//起点到各个点的长度初值为直接和起点连接的长度
    }
    for(int k=1;k<=n;k++)//向最短路径树中加点
    {
        int a=0,min1=inf;
        for(int i=1;i<=n;i++)//找一个目前不在最短路径中的点且离起点最近，即找最小的d[i]
        {
            if(!vis[i]&&d[i]<min1)
            {
                a=i;
                min1=d[i];
            }
        }
        vis[a]=1;//将找到的点加入最短路径树中
        for(int j=1;j<=n;j++)//在通过新加入的a点路径找到离起点更短的路径
        {
            if(!vis[j]&&d[a]+w[a][j]<d[j])
            {
                d[j]=d[a]+w[a][j];
            }
        }
    }
}

int main()
{
    while(scanf("%d%d",&t,&n)==2)
    {
        for(int i=1;i<=n;i++)//先将各个路径初值设为inf
        {
            for(int j=1;j<=n;j++)
            {
                if(i==j)
                {
                    w[i][j]=0;
                }
                else
                {
                    w[i][j]=inf;
                }
            }
        }
        int a,b,c;
        for(int i=0;i<t;i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            if(w[a][b]>c)
            {
                w[a][b]=c;
                w[b][a]=c;
            }
        }
        dijkstra();
        printf("%d\n",d[n]);
    }
    return 0;
}