POJ 【2387】 Til the Cows Come Home

Til the Cows Come Home

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 57350		Accepted: 19500

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

Hint

INPUT DETAILS: 

There are five landmarks. 

OUTPUT DETAILS: 

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

算是写一道题做模板参考吧：

#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
#include<vector>
#define INF 0x3f3f3f3f
#define MAXN 100010
using namespace std;
int dis[MAXN];
struct Pair
{
	int first,second;		//first存点，second存距离 
	bool friend operator < (Pair a,Pair b)
	{
		return a.second > b.second;
	}
}pr,ne;
vector<int > edge[MAXN];//存与之相邻的两个边
int length[1005][1005];//存相邻边的长度 
void dijkstra(){
	memset(dis,INF,sizeof(dis));
	bool vis[MAXN];
	memset(vis,false,sizeof(vis));
	dis[1]=0;
	pr.first=1;
	pr.second=0;
	priority_queue<Pair> Q;
	Q.push(pr);
	while(!Q.empty())
	{
		pr=Q.top();
		Q.pop();
		if(vis[pr.first])
		{
			continue;
		}
		vis[pr.first]=true;
		for(int i=0;i<edge[pr.first].size();i++)
		{
			ne.first=edge[pr.first][i];
			ne.second=pr.second+length[pr.first][ne.first];
			if(ne.second<dis[ne.first])
			{
				dis[ne.first]=ne.second;
				Q.push(ne);
			}
		}	
	}
}
int main()
{
	memset(length,-1,sizeof(length));
	int T,N;
	while(scanf("%d%d",&T,&N)!=EOF)
	{
		for(int i=0;i<T;i++)
		{
			int x,y,z;
			scanf("%d %d %d",&x,&y,&z);
			edge[x].push_back(y);
			edge[y].push_back(x);
			if(length[x][y]==-1)
			length[x][y]=length[y][x]=z;
			else
			length[x][y]=length[y][x]=min(z,length[y][x]);
		}
		dijkstra();
		printf ("%d\n",dis[N]);
	}
	return 0;
 }