POJ-2387 Til the Cows Come Home

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:
Bessie can get home by following trails 4, 3, 2, and 1.

最短路问题（注意的是这是双向牛道）
单源最短路问题（dijkstra朴素版）

#include<iostream>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std; 
typedef pair<int,int>PII;
const int N=1005;
int t,n,a,b,c;
int g[N][N];
int dist[N*N];//记录每个点到1号点的距离
bool st[N]; //每个点的最短路有没有被确定
void dijkstra()
{
	memset(dist,0x3f,sizeof(dist)); //初始化
	dist[1]=0; //编号1的距离为0
	for(int i=0;i<n-1;i++)
	{
		int t=-1; //表示还未确定的
		for(int j=1;j<=n;j++) //遍历所有点
		if(!st[j]&&(t==-1||dist[t]>dist[j])) //如果当前这个点没有被确定最短路，还未确定，当前t不是最短的
					t=j; 
		st[t]=true; //标记一下t的最短路已被确定
		for(int j=1;j<=n;j++)//用t来更新其他点的距离
			dist[j]=min(dist[j],dist[t]+g[t][j]);//用1-t，t-j的长度来更新1-j的长度。
	}
	printf("%d\n",dist[n]);
	
}
int main()
{
	while(~scanf("%d %d",&t,&n))
	{
		memset(g,0x3f,sizeof(g));
		for(int i=0;i<t;i++)
		{
			cin>>a>>b>>c;
			g[a][b]=g[b][a]=min(g[a][b],c);  //具有重边 
		}
		dijkstra();
	}
	return 0;
 }