POJ - 2387-Til the Cows Come Home

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在Farmer John的场地上，Bessie需要找到从草地返回谷仓的最短路径，以确保她能及时休息。场地上有N个地标，通过T条双向牛径相连，每条路径都有特定的长度。本篇详细介绍了如何计算Bessie从地标N到地标1（谷仓）的最短距离。

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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.

#include<iostream>
using namespace std;

int t, n;
struct node
{
	int a, b, c;
}keda[2005];
long long int a[2005];
void find()
{
	int i, j, k;
	a[1] = 0;
	for (i = 2; i <= n; i++)a[i] = 9999999;
	for (i = 1; i <= n; i++)
		for (j = 1; j <= t;j++)
		{
			if (a[keda[j].a] > a[keda[j].b] + keda[j].c) a[keda[j].a] = a[keda[j].b] + keda[j].c;
			if (a[keda[j].b] > a[keda[j].a] + keda[j].c) a[keda[j].b] = a[keda[j].a] + keda[j].c;
		}

}

int main()
{
	int i;
	while (cin >> t >> n)
	{
		for (i = 1; i <= t; i++)cin >> keda[i].a >> keda[i].b >> keda[i].c;
		find();
		cout << a[n] << endl;
	}
	
}