【アルゴリズム】最短経路--poj 2387乳牛帰宅

Til the Cows Come Home
Time Limit: 1000MS
 
Memory Limit: 65536K
Total Submissions: 67017
 
Accepted: 22536
Description
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

テーマ分析:
この問題は単一ソースの最短パス問題でありdijkstraアルゴリズムで解決できる.
コード:

// POJ2387  ---- Til the Cows Come Home
//      --       
//           ，      

#include
#include
#include
#define INT_MAX 0x7fffffff
#define Alone   -1;
using namespace std;


/*
function: Dijkstra(int v,int n,vector &dist,vector &pre,vector> &d)
param:	int v：    
		int n:    
		dist :   
		pre :     
		d  :    
return: void 

  ：1.   0   1  ？    n/ s(n)/s(n+1)
	  2.         
		 
*/

void Dijkstra(int v,int n,vector &dist,vector &pre,vector > &d)
{
	vector s(n+1); //            0-   
	for(int i=1;i<=n;i++)  //    dist，pre 
	{
		dist[i]=d[v][i];
		if(dist[i] < INT_MAX)
		{
			pre[i]=v;
		}
		else pre[i]=Alone; //      
	}
	dist[v]=0; //         0 
	s[v]=true;  //        
	
	for(int i=2;i<=n;i++)  // n-1    
	{
		int best=v;
		int temp=INT_MAX; 
		for(int j=2;j<=n;j++) //       
		{
			if(!s[j] && dist[j]k   (  ) 
			{ 
				int new_dist=dist[best]+d[best][k];
				if(new_dist < dist[k])
				{
					dist[k]=new_dist;
					pre[k]=best;
				}	
			}
		} 
	}  
}

int main()
{
	//freopen("2.txt","r",stdin);
	int n,m;  //n   ，m   
	cin>>m>>n;
	vector > d(n+1,vector(n+1)) ; //     （n*n） 
	//vector >      > >       
	//[Error] '>>' should be '> >' within a nested template argument list
	
	//     
	for(int i=1;i<=n;i++)  //   0  ,    
	{
		for(int j=1;j<=n;j++)
		{
			d[i][j]=INT_MAX; 
		}
	} 
	
	int p,q,value; // p->q
	for(int i=0;i>p>>q>>value;
		if(d[p][q]>value) 
		{
		d[p][q]=value;
		d[q][p]=value; //       
		}
	}
   

	vector  dist(n+1),pre(n+1);
	Dijkstra(1,n,dist,pre,d);

   // cout<