最短路练习1 poj 2387 Til the Cows Come Home

                   题目链接：http://poj.org/problem?id=2387

Til the Cows Come Home

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 51700		Accepted: 17474

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.

最基础的最短路模板，就有一个坑点；思路：更新i点到1点的距离用dis【i】记录；

AC代码：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<map>
#include<string>
#define LL long long
using namespace std;
const int mod = 1e7+7;
const int inf = 0x3f3f3f3f;
const int maxn = 1e6 +10;
int dis[1005];//i点到1点的距离用dis【i】记录；
int v[1005];//标记是否已经最短；
int ma[1005][1005];//存两点的距离；
int m,n,a,b,c;
int main()
{
    while(~scanf("%d%d",&m,&n))
    {
        memset(ma,inf,sizeof(ma));//只有0，-1，inf=0x3f3f3f3f才可以用memset函数初始化，其他的值会乱码
        memset(v,0,sizeof(v));
        for(int i=0; i<m; i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            if(ma[a][b]>c)//《没有这一行是错的》
            ma[a][b]=ma[b][a]=c;
        }
        dis[n]=0;//这里我都弄反了，让求n到1的，我都写成1到n了，不过不影响结果
        v[n]=1;
        for(int i=1; i<n; i++)
            dis[i]=ma[i][n];
        int h=n-1;
        while(h--)
        {
            int sum=inf,point=-1;
            for(int i=1; i<n; i++)//选一个到原点最短的点（未被标记的点）；
                if(!v[i]&&dis[i]<sum)
                {
                    sum=dis[i];
                    point=i;
                }
            if(point !=-1)
            {
                v[point]=1;
                for(int i=1;i<n;i++)//用这个点更新其他点
                    if(!v[i]&&dis[point]+ma[i][point]<dis[i])
                        dis[i]=dis[point]+ma[i][point];
            }
        }
        printf("%d\n",dis[1]);
    }
}