Til the Cows Come Home（最短路，注意重边）

题目链接：http://poj.org/problem?id=2387

 

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

这道题是求最短路问题，点到点，需要注意的是可能有重边，要比较一下：

很明显，我感觉用邻接矩阵更简单一点，临界表感觉好麻烦。。我看了一下旁边的，他邻接表真的有点麻烦。。。

#include
#include
#include

#include
#include
#include
#include
#include
using namespace std;

#define ll long long
#define da    10000000
#define xiao -10000000
#define clean(a,b) memset(a,b,sizeof(a))


int shuzu[1005][1005];
bool biaoji[1005];
int shuaxin[1005];
int n,t; 

void prime(int x)
{
	int i,j;
	for(i=0;i<1005;++i)
		shuaxin[i]=da;
	shuaxin[x]=0;
	for(i=1;i<=n;++i)
	{
		int can=-1,min=da;
		for(j=1;j<=n;++j)
		{
			if(biaoji[j]==0&&shuaxin[j]l)					//如果同样的路径，但距离更短，使用短的 
		{
			shuzu[e][s]=l;					//双向连通图 
			shuzu[s][e]=l;
		}
	}
	prime(1);								//从第0个开始 
	printf("%d\n",shuaxin[n]);				//输出点到点的距离 
	return 0;
}